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.

Let p(-x) = x⁹- 5x⁵ + 4x⁴ + 2x² + 1 = 0
P(x) has only one sign change. It has atmost one positive roots.
Also p(-x) = (-x)⁹ - 5(-x)⁵ + 4(-x)4 +2(-x)² +1 = 0
p(-x) = -x⁹+ 5x⁵+4x⁴+ 2x²+1 = 0
It has only one sign change.
It has atmost one negative roots.
Clearly 0 is not a root.
So maximum number of real roots is 3 and hence there are atleast six imaginary solutions.

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.

Given 2i + 3 is a root
∴ Its conjugate 3 - 2i is also a root of the polynomial equation.
∴ Sum of the roots 3 + 2i + 3 - 2i = 6
Product of the roots = (3 + 2i) (3 - 2i)
= 3² + 2² = 9+ 4 = 13
∴ The polynomial equation of minimum degree with rational co-efficients is
x² - x (sum of the roots) + product of the roots = 0
⇒x² -x (6) + 13 = 0
⇒ x²- 6x + 13 = 0

Since \(2+i\sqrt { 3 } \) is a root of the polynomial equation, its conjugate 2-i\(\sqrt3\) is also a root of the equation:
∴ Sum of the roots \(=2+i\sqrt { 3 } +2-i\sqrt { 3 } =4\)
Product of the roots \(=(2+i\sqrt { 3 } )(2-i\sqrt { 3 } )\)
\(={ 2 }^{ 2 }+{ (\sqrt { 3 } })^{ 2 }\)
\([\because (a+ib)(a-ib)={ a }^{ 2 }+{ b }^{ 2 }]\)
= 4 + 3 = 7
Hence, the polynomial equation of minimum degree with rational co-efficients is
x² - x (sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x(4)+7=0\)
\(\Rightarrow { x }^{ 2 }-4x+7=0\)

Since 2-\(\sqrt{3}\)i is a root of the required polynomial equation with real coefficients, 2 +\(\sqrt{3}\)i is also a root. Hence the sum of the roots is 4 and the product of the roots is 7. Thus x²-4x + 7= 0 is the required monic polynomial equation.

Given that the height of the tree is 12m.
Let x m be the standing part and (12 -x)m be the broken part.
Given \(x=\sqrt [ 3 ]{ 12-x } \)
\(\Rightarrow x=(12-x)^{ \frac { 1 }{ 3 } }\)


Taking power 3 both sides, we get 
⇒ x³ = 12-x
⇒ x³+ x -12 = 0
which is the required mathematical problem.

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.

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) = x⁷-6x⁶+7x⁵+5x²+2x+2
It has only one change of sign. Now,
p(-x) = (-x)⁷ -6(-x)⁶ +7(-x)⁵ +5(-x)² +2(-x)+2
= -x⁷ -6x⁶ -7x⁵ + 5x² - 2x + 2
It has two, change of sign.
Hence, p(x) has one positive root and has at least two negative roots.

We have \(x=\sqrt { 2+x } \)
\(\Rightarrow { x }^{ 2 }=2+x \Rightarrow { x }^{ 2 }-x-2=0\)
\(\Rightarrow x=\frac { 1\pm \sqrt { 1+8 } }{ 2 } \Rightarrow x=\frac { 1\pm 3 }{ 2 } \)
\(\Rightarrow x=\frac { 1+3 }{ 2 } ,\frac { 1-3 }{ 2 } \Rightarrow x=2,-1\)
Also x>0, we get x = 2

The quadratic equation 3x² + 2(a² + 1)x + (a² - 3a + 2)
Will have two roots of opposite sign if it has real roots and the product of the roots is negative.
⇒ 4(a²+1)²-12(a²-3a+2)\(\ge\) 0 and \(\frac { { a }^{ 2 }-3a+2 }{ 3 } <0\)
Both of these conditions are true if 
⇒ a²-3a+2< 0
⇒ (a-1)(a-2)< 0
⇒ 1<a<2

Let ∝, β are the roots of the equation
Sum of the roots \(\alpha +\beta =\frac { -b }{ a } \)
 \(=\frac { [-(a-2)] }{ 1 } =a-2\)
Product of the roots \(=\alpha \beta =\frac { c }{ a } \)
\(=\frac { -(a+1) }{ 1 } =-(a+1)\)
we have \({ \alpha }^{ 2 }{ \beta }^{ 2 }=({ \alpha +\beta ) }^{ 2 }-2\alpha \beta \)
\(={ (a-2) }^{ 2 }+2(a+1)\)
\(={ a }^{ 2 }-4a+4+2a+2\)
\(=(a-1{ ) }^{ 2 }+5\)
Thus \(\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }\) is least if a = 1

Sum of the roots = sin ∝ + cos ∝ = \(\frac{-b}{a}\)
Product of the roots = sin ∝ cos ∝ = \(\frac{c}{a}\)
Now 1 = cos²∝ + sin² ∝
= (sin ∝ +cos ∝)² - 2 sin ∝ cos ∝
\(1=\frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } \Rightarrow 1=\frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \)
⇒ a² = b² - 2ac ⇒ a² + 2ac = b²
Adding c² both sides, a² +2ac+c² = b²+c²
⇒ (a+c)² = b² + c²