Since α, β, and γ are the roots of the equation x³+ px²+ qx + r = 0, we have
Σ₁ α + β + γ = -p and Σ₃ αβγ = -r
\(\Sigma \frac { 1 }{ \beta \gamma } =\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } +\frac { 1 }{ \alpha \beta } =\frac { \alpha +\beta +\gamma }{ \alpha \beta \gamma } =\frac { -p }{ -r } =\frac { p }{ r } \).

Since α and β are the roots of the quadratic equation, we have α + β =\(\frac { 7 }{ 2 } \) and αβ = \(\frac { 13 }{ 2 } \).
Thus, to construct a new quadratic equation,
Sum of the roots = α²+β² = (α+β)²-2αβ =\(\frac { -3 }{ 4 } \)
Product of the roots = α²β² = (αβ)² = \(\frac { 169 }{ 4 }\)
Thus a required quadratic equation is  x²+\(\frac { 3 }{ 4 } x+\frac { 169 }{ 4 } \)= 0. 
From this we see that 4x²+3x+169 = 0 is a quadratic equation with roots α² and β².

Since α and β are the roots of 17x²+ 43x −73 = 0 , we have α + β =\(\frac { -43 }{ 17 } \) and  αβ =\(\frac { -73 }{ 17 } \).
We wish to construct a quadratic equation with roots α + 2 and  β + 2. Thus, to construct such a quadratic equation, calculate
the sum of the roots =  α + β + 4 = \(\frac { -4 }{ 17 } +4=\frac { 25 }{ 17 } \) and
the product of the roots = αβ + 2(α+β)+4 = \(\frac { -73 }{ 17 } +2\left( \frac { -43 }{ 17 } \right) +4=\frac { -91 }{ 17 } \)
Hence a quadratic equation with required roots is x²-\(\frac { 25 }{ 17 } x-\frac { 91 }{ 17 } \) = 0
Multiplying this equation by 17, gives 17x²−25x−91 = 0
which is also a quadratic equation having roots α + 2 and β + 2

The theorem states that the number of positive roots of a polynomial P(x) cannot be more than the number of sign changes in coefficients of P(x). Further it says that the difference between the number of sign changes in coefficients of P(x) and the number of positive roots of the polynomial P(x) is even.
As a negative zero of P(x) is a positive zero of P(−x) we may use the theorem and conclude that the number of negative zeros of the polynomial P(x) cannot be more than the number of sign changes in coefficients of P(−x) and the difference between the number of sign changes in coefficients of P(−x)and the number of negative zeros of the polynomial P(x) is even.
As the multiplication of a polynomial by x^k, for some positive integer k, neither changes the number of positive zeros of the polynomial nor the number of sign changes in coefficients, we need not worry about the constant term of the polynomial. Some authors assume further that the constant term of the polynomial must be non zero.
We note that nothing is stated about 0 as a root, in Descartes rule. But from the very sight of the polynomial written in the customary form, one can say whether 0 is a root of the polynomial or not. Now let us verify Descartes rule by means of certain polynomials

Consider the polynomial equation
\(a_{n} x^{n}+a_{n-1} x^{n-1}+a_{n-2} x^{n-2}+\cdots+a_{2} x^{2}+a_{1} x+a_{0}=0, \)  ......(1)
Replacing \(x \text { by } \frac{1}{x}\) in (1), we get
\(P\left(\frac{1}{x}\right)=\frac{a_{n}}{x^{n}}+\frac{a_{n-1}}{x^{n-1}}+\frac{a_{n-2}}{x^{n-2}}+\cdots+\frac{a_{2}}{x^{2}}+\frac{a_{1}}{x}+a_{0}=0\)    .....(2)
Multiplying both sides of (2) by xⁿ, we get
\(x^{n} P\left(\frac{1}{x}\right)=a_{0} x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\cdots+a_{n-2} x^{2}+a_{n-1} x+a_{n}=0\)       .......(3)
Now, (1) is a reciprocal equation \(\Leftrightarrow P(x)=\pm x^{n} P\left(\frac{1}{x}\right) \Leftrightarrow\) (1) and (3) are same
This is possible \(\Leftrightarrow \frac{a_{n}}{a_{0}}=\frac{a_{n-1}}{a_{1}}=\frac{a_{n-2}}{a_{2}}=\cdots=\frac{a_{2}}{a_{n-2}}=\frac{a_{1}}{a_{n-1}}=\frac{a_{0}}{a_{n}} .\)
Let the proportion be equal to λ. Then, we get \(\frac{a_{n}}{a_{0}}=\lambda \ \text { and } \ \frac{a_{0}}{a_{n}}=\lambda\) Multiplying these equations, we get \(\lambda\)² = 1. So, we get two cases \(\lambda\) = 1and \(\lambda\) = −1.
Case (i) :
\(\lambda\) = 1 In this case, we have \(a_{n}=a_{0}, a_{n-1}=a_{1}, a_{n-2}=a_{2}, \cdots\)
That is, the coefficients of (1) from the beginning are equal to the coefficients from the end.
Case (ii) :
\(\lambda\) = −1 In this case, we have \(a_{n}=a_{0}, a_{n-1}=a_{1}, a_{n-2}=a_{2}, \cdots\)
That is, the coefficients of (1) from the beginning are equal in magnitude to the coefficients from the end, but opposite in sign.

When a_n = 1, if there is a rational root \(\frac{p}{q}\) then as per theorem 3.5 q is a factor of a_n , then we must have q = ±1. Thus p must be an integer. So a monic polynomial with integer coefficient cannot have non-integral rational roots. So when a_n = 1, if at all there is a rational root, it must be an integer and the integer should divide a₀. (We say an integer a divides an integer b, if b = ad for some integer d.)
As an example let us consider the equation x² − 5x − 6 = 0. The divisors of 6 are ±1, ± 2, ± 3, ± 6. From Rational Root Theorem, we can conclude that ±1, ± 2, ± 3, ± 6 are the only possible solutions of the equation. It does not mean that all of them are solutions. The two values −1 and 6 satisfy the equation and other values do not satisfy the equation.
Moreover, if we consider the equation x² + 4 = 0, according to the Rational Root theorem, the possible solutions are ±1, ± 2, ± 4; but none of them is a solution. The Rational Root Theorem helps us only to guess a solution and it does not give a solution.

We prove the theorem by assuming that the quadratic equation is a monic polynomial equation. The result for non-monic polynomial equation can be proved in a similar way.
Let p and q be rational numbers such that \(\sqrt{q}\) is irrational. Let p + \(\sqrt{q}\) be a root of the equation x² + bx + c = 0 where b and c are rational numbers.
Let α be the other root. Computing the sum of the roots, we get
α + p + \(\sqrt{q}\) = −b
and hence \(\alpha+\sqrt{q}=-b-p \in \mathbb{Q}\). Taking − b − p as s, we have α + \(\sqrt{q}\)  = s
This implies that
α = s −\(\sqrt{q}\)
Computing the product of the roots, we get
(s − \(\sqrt{q}\))( p + \(\sqrt{q}\)) = c
and hence \((s p-q)+(s-p) \sqrt{q}=c \in \mathbb{Q}\) . Thus s − p = 0 . This implies that s = p and hence we get α = p − \(\sqrt{q}\) . So, the other root is p − \(\sqrt{q}\) .

Let \(P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{o}=0\) be a polynomial equation with real coefficients
Let z₀ be a root of this polynomial equation So, P( z₀ ) = 0. Now
\( P\left(\overline{z_{0}}\right) =a_{n} \bar{z}_{0}^{n}+a_{n-1} \bar{z}_{0}^{n-1}+\cdots+a_{1} \bar{z}_{0}+a_{0} \)
\(=a_{n} \overline{z_{0}{ }^{n}}+a_{n-1} \overline{z_{0}{ }^{n-1}}+\cdots+a_{1} \overline{z_{0}}+a_{0} \)
\(=\overline{a_{n}} \overline{z_{0}{ }^{n}}+\overline{a_{n-1}} \overline{z_{0}{ }^{n-1}}+\cdots+\overline{a_{1}} \overline{z_{0}}+\overline{a_{0}} \ \left(a_{r}=\overline{a_{r}} \text { as } a_{r} \text { is real for all } r\right) \)
\(=\overline{a_{n} z_{0}{ }^{n}}+\overline{a_{n-1} z_{0}{ }^{n-1}}+\cdots+\overline{a_{1} z_{0}}+\overline{a_{0}}\)
\(=\overline{a_{n} z_{0}{ }^{n}+a_{n-1} z_{0}{ }^{n-1}+\cdots+a_{1} z_{0}+a_{0}}=\overline{P\left(z_{0}\right)}=\overline{0}=0\)
That is P(\(\bar{z}_{0}\)) = 0 this implies that whenever z₀ is a root (i.e. P( z₀ ) = 0), its conjugate is \(\bar{z}_{0}\) also a root.

Using this, we can prove that a polynomial equation of degree n has at least n roots in C when the roots are counted with their multiplicities. This statement together with our discussion above says that a polynomial equation of degree n has exactly n roots in C when the roots are counted with their multiplicities. Some authors state this statement as the fundamental theorem of algebra.

Let p(x) = x⁹ + 9x⁷ + 7x⁵ + 5x³ + 3x
p(x) has no sign change.
p(-x)⇒ (-x)⁹ + 9(-x)⁷ + 7(-x)⁵ + 5(-x)³ + 3(-x)
= -x⁹ - 9x⁷ - 7x⁵- 5x³ -3x
p(-x) also has no sign change
∴ p(x) has no positive and no negative root.

Let p(x) = x⁹ -5x² - 14x⁷ = 0
p(x) has only one sign change
Also p(-x) = (-x)⁹ - 5 (-x)⁸ - 14(-x)⁷ = 0
⇒ p(-x) = -x⁹ -5x⁸ + 14x⁷ = 0
p(-x) has only one sign change
∴ p(-x) has at most one positive and one negative root.

Clearly there are 2 sign changes for the given polynomial P(x) and hence number of positive roots of P(x) cannot be more than two. Further, as P(-x) = -9x⁹- 2x⁵- x⁴- 7x²+ 2, there is one sign change for P(-x) and hence the number of negative roots cannot be more than one. Clearly 0 is not a root. So maximum number of real roots is 3 and hence there are atleast six imaginary roots.

sin²x - 5 sinx + 4 = 0
put y = sin x
⇒ y²-5y+4 = 0
⇒ (y-4)(y-1) = 0
⇒ y = 4, 1
Case(i) 
When y = 4, sin x = 4 and no solution for sin x = 4 since the range the sin function is [-1, 1]
Case (ii)
When y = 1, sin x = 1
⇒ sin x = sin \(\frac{\pi}{2}\) [\(\because sin \frac {\pi}{2}=1\)]
\(x=n \pi+(-1)^{n} \frac{\pi}{2} \forall n \in z\).

2cos²x - 9cosx + 4 = 0 ............ (1)
The left hand side of this equation is not a polynomial in x. But it looks like a polynomial. In fact, we can say that this is a polynomial in cos x. However, we can solve the equation (1) by using our knowledge on polynomial equations. If we replace cos x by y, then we get the polynomial equation 2y²- 9y + 4 = 0 for which 4 and \(\frac{1}{2}\) are solutions.
From this we conclude that x must satisfy cos x = 4 or cos x = \(\frac{1}{2}\).
But cos x = 4 is never possible, if we take cos x = \(\frac{1}{2}\), then we get infinitely many real numbers x satisfying cos x = \(\frac{1}{2}\); in fact, for all n\(\in \)Z, x = 2nπ ±\(\frac { \pi }{ 3 } \) are solutions for the given equation (1).
If we repeat the steps by taking the equation cos²x - 9 cosx + 20 = 0, we observe that this equation has no solution.

If P(x) denotes the polynomial in the equation, then P(2) = 0.
Hence 2 is a root of the polynomial.
To find other roots, we divide the given polynomial x³−5x²−4x + 20 by x − 2 and get Q(x) = x² −3x−10 as the quotient.
Solving Q(x) = 0 we get −2 and 5 as roots.
Thus 2, −2, 5 are the solutions of the given equation.

Since \(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) is a root, x-\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) is a factor. To remove the outermost square root, we take x +\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) as another factor and find their product.
\(\left( x+\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \right) \left( x-\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \right) ={ x }^{ 2 }-\frac { \sqrt { 2 } }{ \sqrt { 3 } } \)
Still we didn’t achieve our goal. So we include another factor x²+\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) and get the product.
\(\left( { x }^{ 2 }-\frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) \left( { x }^{ 2 }+\frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) ={ x }^{ 4 }-\frac { 2 }{ 3 } \)
So, 3x⁴- 2 = 0 is a required polynomial equation with the integer coefficients.
Now we identify the nature of roots of the given equation without solving the equation. The idea comes from the negativity, equality to 0, positivity of Δ = b²- 4ac.

Let α, β, γ and δ be the roots of ax⁴+ bx³+ cx²+ dx + e = 0
Σ₁ = α + β + γ + δ = -\(\frac { b }{ a } \),
Σ₂ = αβ + αγ + αδ + βγ + βδ + γδ = \(\frac { c }{ a } \),
Σ₃ = αβγ + αβδ +  αγδ +  βγδ = -\(\frac { d }{ a } \)
Σ₄ = αβγδ =\(\frac { e }{ a } \)
We have to find α² + β² + γ² + δ² 
Applying the algebraic identity
(a+b+c+d)² ≡ a²+b²+c²+d²+2(ab+ac+ad+bc+bd+cd),
we get
α² + β² + γ² + δ² = (α + β + γ + δ)²-2(αβ + αγ + αδ + βγ + βδ + γδ)
= \(\left( \frac { b }{ a } \right) ^{ 2 }-2\left( \frac { c }{ a } \right) \)
= \(\frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \).

The length and breadth of the cuboid are x + 1.
x + 2 and x + 3
[∵ they are increased by 1, 2, 3 units]
Also volume = V + 52 .
[since V is increased by 52]
∴ V + 52 = (x +1)(x + 2)(x + 3) .........(1)
⇒ V = (x + 1) (x + 2) (x + 3) - 52
Here a = -1, β = -2, ૪ = -3
⇒ V = x³-x²(α+β+૪)+x(αβ+β૪+૪α)-αβ૪ = 52
⇒ V = x³-x²(-1-2-3)+x(2+6+3)-(-1)(-2)(-3) = 52
⇒ V = x³-x²(-6)+x(11)+6-52
⇒ x³ = x²+6x²+11x+6-52
⇒ 6x²+11x-46 = 0
⇒ (6x+23)(x-2) = 0
⇒ (6x+23)(x-2) = 0
⇒ x = 2

∴ Volume of the cube = x³ = 2³ = 8.
Volume of a cuboid = 52 + 8 = 60
[∵ x = \(\frac{-23}{6}\) is not possible as x represents the side of the cube]

Let P(x) be the polynomial under consideration.
The number of sign changes for P(x) and P(−x) are 2 and 1 respectively. Hence it has at most two positive roots and at most one negative root. Since the difference between number of sign changes in coefficients of P(−x) and the number of negative roots is even, we cannot have zero negative roots. So the number of negative roots is 1. Since the difference between number of sign changes in coefficient of P(x) and the number of positive roots must be even, we must have either zero or two positive roots. But as the sum of the coefficients is zero, 1 is a root. Thus we must have two and only two positive roots Obviously the other two roots are imaginary numbers.

x⁸- 3x + 1 = 0
Here a_n = 1, a_o = 1
If \(\frac{p}{q}\) is a root of the polynomial, then as 
(p, q) = 1p is a factor of a_o = 1 and q is a factor of a_n = 1
Since 1 has no factors, the given equation has no rational roots.

12x + 8x = 29x²- 4
This equation can be re-written as
12x³ - 29x² + 8x + 4 = 0

∴ x = 2 is a root and the remaining factor is 
12x²- 5x - 2
⇒ (3x-2)(4x+1) = 0
⇒ 3x-2 = 0 or 4x+1 = 0

⇒ x = \(\frac{2}{3}\)
x = \(\frac{-1}{4}\)
∴ The roots are 2, \(\frac{2}{3}\), \(\frac{-1}{4}\)

Let f(x) = 8x³- 2x² - 7x +3 = 0
Here sum of the co-efficients of odd terms = 8 - 7 = 1
and sum of the co-efficients of even terms = -2 +3 = 1
Hence, x = - 1 is a root of f(x)
Let us divide f(x) by (x + 1)

∴ The other factor is 8x² - 10x + 3
\(\Rightarrow x=\frac { 10\pm \sqrt { 100-4(8)(3) } }{ 2\times 8 } \)
\(\Rightarrow x=\frac { 10\pm \sqrt { 100-96 } }{ 16 } \Rightarrow x=\frac { 10\pm 2 }{ 16 } \)
\(\Rightarrow x=\frac { 12 }{ 16 } \ or \)
\(x=\frac { 8 }{ 16 } \Rightarrow x=\frac { 3 }{ 4 } ,\frac { 1 }{ 2 } \)
∴ The roots are -1, \(\frac{1}{2},\frac{3}{4}\)

x = 1 
y = x² -5x + 6
y = 1 -5 + 6 = 2
x = 2
y = 4 -10 + 6 = 0
x = 0 
y = 6
x = 3
y = 9 -15 + 6 = 12
x = -1 
y = 1 + 5 + 6 = 12
x = 4
y = 16 - 20 + 6 = 2

(1, 2), (0, 6), (-1, 12)
P(x) = (x²-5x + 6) (x²-5x+16)
= x⁴- 5x³+16x²-5x+25x²- 80x + 6x²- 30x + 96 = 0
x⁴-10x³+ 47x² -110x + 96 = 0
It has two sign changes
\(\therefore\) it has two positive real roots
P(-x) = x⁴-10x³+ 47x² -110x + 96
It has no sign changes, no negative real roots
y = x²- 5x + 16

Let p(x) = 9x⁹ - 4x⁸ + 4x⁷ - 3x⁶ + 2x⁵ + x3 + 7x2 + 7x + 2 = 0
Clearly there are 4 sign changes for the given| polynomial P(x) and hence number of positive roots of P(x) can't be more than four.
hence the number of positive roots of p(x) cannot be more than 4.
p(-x) = 9(-x)⁹ - 4(-x)⁸ + 4(-x)⁷ - 3(-x)⁶+ 2(-x)⁵ + (-x)³ + 7 (-x)² + 7(-x) + 2
There are two sign changes. Hence the number of negative roots can't be more than two.
It has atmost 4 positive roots and atmnost two negative roots.

4^x- 3(2^{x + 2}) + 2⁵ = 0
\(\Rightarrow \left( { 2 }^{ 2 } \right) ^{ x }-3({ 2 }^{ x })({ 2 }^{ 2 })+{ 2 }^{ 5 }=0\)
\(\left( { 2 }^{ x } \right) ^{ 2 }-3\left( { 2 }^{ x } \right) ({ 2 }^{ 2 })=0\)
\(\left( { 2 }^{ x } \right) ^{ 2 }-12\left( { 2 }^{ x } \right) +32=0\)
Put \({ 2 }^{ x }=y\)
\({ y }^{ 2 }-12y+32=0\)
(y - 8)(y - 4) = 0
y = 8, 4 
Case (i) when \(y=8,{ 2 }^{ x }=8\Rightarrow { 2 }^{ x }={ 2 }^{ 3 }\Rightarrow x=3\)
Case (ii) when \(y=4,{ 2 }^{ x }=4\Rightarrow { 2 }^{ x }={ 2 }^{ 2 }x=\pm 2\)
∴ The roots are 2, 3

Put \(\sqrt { \frac { x }{ a } } =y\Rightarrow 2y+\frac { 3 }{ y } =\frac { b }{ a } +\frac { 6a }{ b } \)
\(\Rightarrow \frac { { 2y }^{ 2 }+3 }{ y } =\frac { { b }^{ 2 }+{ 6a }^{ 2 } }{ ab } \)
\(\Rightarrow ab({ 2y }^{ 2 }+3)=\left( { b }^{ 2 }+{ 6a }^{ 2 } \right) y\)
\(\Rightarrow 2ab{ y }^{ 2 }+3ab-\left( { b }^{ 2 }+{ 6a }^{ 2 } \right) =0\)
\(\Rightarrow 2ab{ y }^{ 2 }-y\left( { b }^{ 2 }+{ 6a }^{ 2 } \right) +3ab=0\)
\(\Rightarrow 2ab{ y }^{ 2 }-{ b }^{ 2 }y-{ 6a }^{ 2 }+3ab=0\)
\(\Rightarrow by(2ay-b)-3a(2ay-b)=0\)
\(\Rightarrow (2ay-b)(by-3a)=0\)
\(\Rightarrow 2ay=b,\ by=3a\)
\(\Rightarrow y=\frac { b }{ 2a } ,y=\frac { 3a }{ b } \)
Case (i) When \(y=\frac { b }{ 2a } \)
\(\Rightarrow \sqrt { \frac { x }{ a } } =\frac { b }{ 2a } \Rightarrow \frac { x }{ a } =\frac { { b }^{ 2 } }{ { 4a }^{ 2 } } \Rightarrow x=\frac { { b }^{ 2 } }{ 4a } \)
Case (ii) When \(y=\frac { 3a }{ b } \)
\(\sqrt { \frac { x }{ a } } =\frac { 3a }{ b } \Rightarrow \frac { x }{ a } =\frac { 9a^{ 2 } }{ { b }^{ 2 } } \Rightarrow x=\frac { { 9a }^{ 3 } }{ { b }^{ 2 } } \)
∴ The roots are \(\frac { { b }^{ 2 } }{ 4a } ,\frac { 9a^{ 3 } }{ b^{ 2 } } \)

Let P(x) be the polynomial under consideration.
The number of sign changes for P(x) and P(−x) are zero and hence it has no positive roots and no negative roots. Clearly zero is not a root. Thus the polynomial has no real roots and hence all roots of the polynomial are imaginary roots.

\(8x^{ \frac { 3 }{ 2x } }-8x^{ \frac { -3 }{ 2x } }\) = 63
\(\Rightarrow 8\left[ { \left( { x }^{ \frac { 1 }{ 2n } } \right) }^{ 3 }-{ \left( { x }^{ \frac { -1 }{ 2n } } \right) }^{ 3 } \right] =63\)
Put \({ x }^{ \frac { 1 }{ 2n } }=y\)
\(\Rightarrow 8\left( { y }^{ 2 }-\frac { 1 }{ { y }^{ 3 } } \right) =63\)
\(\Rightarrow { y }^{ 3 }-\frac { 1 }{ { y }^{ 3 } } =\frac { 63 }{ 8 } \Rightarrow \frac { { y }^{ 6 }-1 }{ { y }^{ 3 } } =\frac { 63 }{ 8 } \)
\(\Rightarrow { 8y }^{ 6 }-8=63{ y }^{ 3 }\)
\(\Rightarrow { 8y }^{ 6 }-{ 63y }^{ 3 }-8=0\)
\(\Rightarrow { 8t }^{ 2 }-63t-8=0\ [where\quad t={ y }^{ 3 }]\)
\(\Rightarrow (8t-1)(t-8)=0\)
\(\Rightarrow t=\frac { 1 }{ 8 } ,8\)
Case (i): when \(t=8,\Rightarrow { y }^{ 3 }=8\Rightarrow { y }^{ 2 }={ 2 }^{ 3 }\)
\(\Rightarrow y=2\)
Case (ii): when \(t=\frac { 1 }{ 8 } ,{ y }^{ 3 }=\frac { 1 }{ 8 } \Rightarrow y=\frac { 1 }{ 2 } \)
When \(y=2,{ x }^{ \frac { 1 }{ 2n } }=2\)
\(\Rightarrow x={ (2 })^{ 2n }\quad \Rightarrow x={ ({ 2 }^{ 2 }) }^{ n }\)
\(\Rightarrow x={ 4 }^{ n }\)
When \(y=\frac { 1 }{ 2 } ,{ x }^{ \frac { 1 }{ 2n } }=\frac { 1 }{ 2 } \Rightarrow x={ \left( \frac { 1 }{ 2 } \right) }^{ 2n }\)
\(\Rightarrow x={ \left( \frac { 1 }{ { 2 }^{ 2 } } \right) }^{ n }=\frac { 1 }{ { 4 }^{ n } } \)
Hence the roots are 4ⁿ.

Since the sum of the co-efficients = 2 - 1- 1 = 0
x = 1 is a root.

∴ x = 1 is a root and the remaining factor is
2x²+x+1
\(\Rightarrow x=\frac {-1\pm \sqrt { { 1 }^{ 2 }-4(2)(1) } }{ 2 } \)
\(\Rightarrow x=\frac { -1\pm \sqrt { -7 } }{ 2 } \) Which is a complex root.
\(\Rightarrow x=\frac { -1\pm \sqrt { -7 } }{ 2 } \)
∴ x = 1 is the rational root.

According to our notations, a_n= 2 and a₀ = 3.
If \(\frac{p}{q}\) is a zero of the polynomial, then as (p, q) = 1, p must divide 3 and q must divide 2.
Clearly, the possible values of p are 1, −1, 3, −3 and the possible values of q are 1, −1, 2, −2.
Using these p and q we can form only the fractions \(\pm \frac { 1 }{ 1 } ,\pm \frac { 1 }{ 2 } ,\pm \frac { 3 }{ 2 } ,\pm \frac { 3 }{ 1 } \). 
Among these eight possibilities, after verifying by substitution, we get \(\frac { -3 }{ 2 } \) is the only rational zero.
To find other roots, we divide the given polynomial 2x³+ 3x²+ 2x + 3 by 2x + 3 and get x²+1 as the quotient with zero remainder. Solving x²+1 = 0, we get i and −i as roots. Thus \(\frac { -3 }{ 2 } \), -i, i are the roots of the given polynomial equation.

Since the sum of the co-efficients is
2 - 9 + 10 - 3 = 12 - 12 = 0
⇒ x = 1 is a root of (x)
∴ (x - 1) is a factor of (x)
To find the other factor, let us divide f(x) by x-1 

[Using synthetic division]
f(x) = (x-1)(x - 3)(2x - 1) = 0
⇒x - 1 = 0, x - 3 = 0 or 2x - 1 = 0
⇒ x = 1, x = 3x, x = \(\frac{1}{2}\)
Hence the roots are 1, 3, \(\frac{1}{2}\).

Given cubic equation is 2x³-6x²+3x+k = 0
Here, a = 2, b = -6, c = 3, d = k
Let ∝, β, ૪ be the roots
Given ∝ = 2(β+૪) ⇒ \(\frac{\alpha}{2}\) = β+૪ ...(1)
Now, \(\alpha +\beta +\gamma =\frac { -b }{ a } =-\frac { (-6) }{ 2 } =3\)
\(\frac { \alpha }{ 2 } +\alpha =3\Rightarrow \frac { \alpha +2\alpha }{ 2 } =3\Rightarrow \frac { 3\alpha }{ 2 } =3\)
\(\Rightarrow \alpha =2\)
\(\alpha \beta \gamma =\frac { -d }{ a } =\frac { -k }{ 2 } \Rightarrow 2.\beta \gamma =\frac { -k }{ 2 } \)
\(\beta \gamma =\frac { -k }{ 4 } ...(2)\)
Also, \(\alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a } \) 
\(2\beta +\beta \gamma +2\gamma =\frac { 3 }{ 2 } \)
\(2(\beta +\gamma )+\beta \gamma =\frac { 3 }{ 2 } \)
\(\alpha \frac { -k }{ 4 } =\frac { 3 }{ 2 } \quad [from(1)\& (2)]\)
Also, \(2-\frac { k }{ 4 } =\frac { 3 }{ 2 } [\because \alpha =2]\)
\(2-\frac { 3 }{ 2 } =\frac { k }{ 4 } \Rightarrow \frac { 1 }{ 2 } =\frac { k }{ 4 } \)
\(\\ k=\frac { 4 }{ 2 } \Rightarrow k=2\)
From(2), \(\beta \gamma =\frac { -k }{ 4 } =\frac { -2 }{ 4 } =\frac { -1 }{ 2 } \)\(\Rightarrow \gamma =\frac { -1 }{ 2\beta } \)
From \((1),\beta +\gamma =\frac { \alpha }{ 2 } =\frac { 2 }{ 2 } =1\)
Substituting \(\gamma =\frac { -1 }{ 2\beta } \) We get
\(\beta -\frac { 1 }{ 2\beta } =1\Rightarrow 2{ \beta }^{ 2 }-1=2\beta \Rightarrow 2\beta -2\beta -1=0\)
\(\beta =\frac { 2\pm \sqrt { 4-4(2)(-1) } }{ 4 } =\frac { 2\pm \sqrt { 4+8 } }{ 4 } \)
\(=\frac { 2\pm \sqrt { 12 } }{ 4 } =\frac { 2\pm 2\sqrt { 3 } }{ 4 } \)
\(\beta =\frac { 1\pm \sqrt { 3 } }{ 2 } \)
Hence the roots are \(2,\frac { 1+\sqrt { 3 } }{ 2 } ,\frac { 1-\sqrt { 3 } }{ 2 } \) and k = 2

Since sum of two of its roots vanishes, let the roots be ∝, -∝ and β 
\(\alpha -\alpha +\beta =\frac { -b }{ a } =\frac { 1 }{ 2 } \)
\(\Rightarrow \ \beta =\frac { 1 }{ 2 } \)
Also, \(\alpha \beta \gamma =\frac { -d }{ a } \)
\(\Rightarrow \alpha (-\alpha )(\frac { 1 }{ 2 } )=\frac { -9 }{ 2 } \)
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ 2 } =\frac { 9 }{ 2 } \Rightarrow { \alpha }^{ 2 }=9\Rightarrow \alpha =\pm 3\)
\(\Rightarrow \alpha =3\)
∴ The roots are 3, -3 and \(\frac{1}{2}\).

Let the roots be in H.P. Then, their reciprocals are in A.P. and roots of the equation
\(\left( \frac { 1 }{ x } \right) ^{ 3 }+p\left( \frac { 1 }{ x } \right) ^{ 2 }+q\left( \frac { 1 }{ x } \right) \)+ r = 0 ⇔ rx³ + qx² + px + 1 = 0.....(1)
Since the roots of (1) are in A.P., we can assume them as α-d, α, α+d
Applying the Vieta’s formula, we get
Σ₁ = (α-d)+α+(α+d) = -\(\frac { q }{ r } \) ⇒ 3α = -\(\frac { q }{ r } \) ⇒ α = -\(\frac { q }{ 3r } \)
But, we note that α is a root of (1). Therefore, we get
\(r\left( -\frac { q }{ 3r } \right) ^{ 2 }+q\left( -\frac { q }{ 3r } \right) ^{ 2 }+p\left( -\frac { q }{ 3r } \right) \) + 1 = 0 ⇒ q³ + 3q³ - 9pqr + 27r² = 0 ⇒ 2q³ + 27r².

We observe that the sum of the coefficients of the odd powers and that of the even powers are equal.
Hence −1 is a root of the equation.
To find other roots, we divide 2x³+11x²-9x-18 by x+1 and get 2x²+9x-18 as the quotient.
Solving this we get \(\frac{3}{2}\) and -6 as roots.
Thus -6, -1, \(\frac{3}{2}\) are the roots or solutions of the given equation.

The sum of the coefficients of the polynomial is 0. Hence 1 is a root of the polynomial. To find other roots, we divide x³- 3x²- 33x + 35 by x-1 and get x²-2x-35 as the quotient. Solving this we get 7 and −5 as roots. Thus 1, 7, −5 form the solution set of the given equation.

The given equation is
x⁴- 9x² + 20 = 0
This is a fourth degree equation. If we replace x² by y then we get the quadratic equation
y²- 9y + 20 = 0
It is easy to see that 4 and 5 as solutions for y²- 9y + 20 = 0. Now taking x² = 4 and x² = 5, we get 2, -2, \(\sqrt{5}\), -\(\sqrt{5}\) as solutions of the given equation.
We note that the technique adopted above can be applied to polynomial equations like x⁶-17x³+30 = 0, ax^2k+ bx^k + c = 0 and in general polynomial equations of the form a_nx^kn + a_n-1x^k(n-1) + .... + a₁x^k + a₀ = 0 where k is any positive integer.

Given equation is 2x² + kx + k = 0
Here a = 2, b = k, c = k
\(\therefore\) Discriminant 
∆ = b²-4ac = k²-4 (2)(k)
= k²-8k = 0
k (k - 8) = 0
Since k is real, the possible values of k are
k < 0, k = 0 or 8
case (i) when 0< k< 8, ∆ < 0
⇒  The roots are imaginary
case (ii) when k = 0 or 8, ∆ = 0,
∴ The roots are real and equal
Case (iii) When k > 8, ∆ = k (k - 8) > 0
⇒ The roots are real and distinct

By choosing the coordinate axes suitably, we take the equation of the circle as x²+ y² = r² and the equation of the straight line as y = mx + c. We know that the points of intersections of the circle and the straight line are the points which satisfy the simultaneous equation
x² + y² = r²
y = mx + c .......... (2)
If we substitute mx + c for y in (1), we get
x²+(mx + c)²- r² = 0
which is same as the quadratic equation
(1+m²)x²+2mcx+(c²- r²) = 0 ............(3)
This equation cannot have more than two solutions, and hence a line and a circle cannot intersect at more than two points.
It is interesting to note that a substitution makes the problem of solving a system of two equations in two variables into a problem of solving a quadratic equation.
Further we note that as the coefficients of the reduced quadratic polynomial are real, either both roots are real or both imaginary. If both roots are imaginary numbers, we conclude that the circle and the straight line do not intersect. In the case of real roots, either they are distinct or multiple roots of the polynomial. If they are distinct, substituting in (2), we get two values for y and hence two points of intersection. If we have equal roots, we say the straight line touches the circle as a tangent. As the polynomial (3) cannot have only one simple real root, a line cannot cut a circle at only one point.

Let  α, β and γ  be the roots of x³+ ax²+ bx + c = 0
Then, we get
Σ₁ = α + β + γ = -a ....(1)
Σ₂ = αβ + βγ + γα = b ...(2)
Σ₃ = αβγ = -c ...(3)
We have to form the equation whose roots are  α², β² and γ².
Using (1), (2) and (3), we find the following
Σ₁ = α² + β² + γ² = (α + β + γ )² - 2( αβ + βγ + γα) = (-a)² -2(b) = a²-2b,
Σ₂ = α²β² +  β²γ² + γ²α² = (αβ + βγ + γα)² - 2((αβ)( βγ)(γα) + (γα)(αβ))
= (αβ + βγ + γα)² - 2αβγ (β + γ + α) = (b)² - 2(-c)(-a) = b²-2ca
Σ₃ = α²β²γ² = (αβγ)² = (-c)² = c²
Hence, the required equation is
x³-(α² + β² + γ²)x² + (α²β² + β²γ² + γ²α²)x -  α²β²γ² = 0
That is, x³-(a²-2b)x² + (b²-2ca)x-c² = 0

Since two roots are in the ratio p : q : r, we can assume the roots as pλ, qλ and rλ .
Then, we get
Σ₁ = pλ + qλ + rλ = -a .....(1)
Σ₂ = (pλ)(qλ)+(qλ)(rλ)+(rλ)(pλ) ..........(2)
Σ₃ = (pλ)(qλ)(rλ) = -c .....(3)
Now, we get
(1) ⇒ λ = -\(\frac { a }{ p+q+r } \) ..........(4)
(3) ⇒ λ³ = \(\frac { c }{ pqr } \) ...........(5)
Substituting (4) in (5), we get
\(\left( \frac { a }{ p+q+r } \right) ^{ 3 }=-\frac { c }{ pqr } \) ⇒ pqra³ = c(p+q+ r)³.

Given equation is 2x⁴- 8x + 6x²- 3 = 0
Here a = 2, b = -8, c = 6, d = 0, e = -3
Let ∝, β, ૪ and \(\delta \) be the roots of equation (1)
Then by Vieta's formula,
\(\sum { _{ 1 }= } \alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -(-8) }{ 2 } =4\)
\(\sum { _{ 2 } } =\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { 6 }{ 2 } =3\)
\(\sum { _{ 3 } } =\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =\frac { 0 }{ a } \)
\(\sum { _{ 4 } } =\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { -3 }{ 2 } \)
Now, (a+b+c+d)² = a²+b²+c²+d²+2(ab+ac+ad+bc+cd)
⇒ n∝²+β²+૪²+\(\delta\)² = (∝ + β + ૪ + \(\delta\))²-2(\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta \))
∝² + β² + ૪² = 4²-2(3) = 16 - 6 = 10

Given cubic equation 3x²-16x²+23x-6 = 0
Let ∝, \(\frac{1}{\alpha}\) and ૪ be the roots of the equation
[∵ product of two roots is 1]
\((1)\rightarrow { x }^{ 2 }-\frac { 16 }{ 3 } { x }^{ 2 }+\frac { 23 }{ 3 } -2=0 \) ......(1)
comparing  (1) with 
\({ x }^{ 3 }-\left( \frac { \alpha +\beta +\gamma }{ \alpha } \right) +\left( \alpha \frac { 1 }{ \alpha } +\frac { 1 }{ \alpha } .\gamma +\gamma \alpha \right) \)
\(-\alpha \frac { 1 }{ \alpha } .\gamma =0\) ..........(2)
We get, 
\(\alpha +\frac { 1 }{ \alpha } +\gamma =\frac { 16 }{ 3 } \) .......(3)
\(1+\frac { \gamma }{ \alpha } +\gamma \alpha =\frac { 23 }{ 3 } \)
\(\alpha .\frac { 1 }{ \alpha } .\gamma =2\Rightarrow \gamma =2\) ............(4)
Substituting ૪ = 2 in (3)
 \(\alpha +\frac { 1 }{ \alpha } +2=\frac { 16 }{ 3 } \)
\(\Rightarrow \alpha +\frac { 1 }{ \alpha } =\frac { 16 }{ 3 } -2=\frac { 16-6 }{ 3 } =\frac { 10 }{ 3 } \)
\(\Rightarrow \frac { { \alpha }^{ 2 }+1 }{ \alpha } =\frac { 10 }{ 3 } \)

\({ 3x }^{ 2 }+3=10\alpha \)
\({ 3\alpha }^{ 2 }-10\alpha +3=0\)
\(\alpha =\frac { -10 }{ 3 } or\ \alpha =\frac { 1 }{ 3 } \)
\((3\alpha +10)(3\alpha -1)=0\)
\(\alpha =\frac { -10 }{ 3 } \) is not possible \(\Rightarrow \alpha =\frac { 1 }{ 3 } \)
[\(\because \alpha =\frac { -10 }{ 3 } \) will not satisfy(5)]
∴ The roots are 3, \(\frac{1}{3}\), 2.