Ananya had 4x red, 5x blue and 3x yellow marbles. Let number of blue marbles given to Neha be y. According to question
4x+ y= 20
and \(\frac{5x - y}{3x} = \frac{7}{9}\)
Number of red marbles = 12

Given, the ones digit is x and the tens digit is y. According to CEO's clue,
y - 2x = 1..(i)
According to VP's clue,
y + 3 = 3x ..(ii)
Now, solve Eqs. (i) and (ii).
Required passcode is 4994

Given equations are
\(\frac{3 x-4 y}{2}=10 \Rightarrow 3 x-4 y=20\) 
and \(\frac{3 x+2 y}{4}=2 \Rightarrow 3 x+2 y=8\) 
x = 4, y = - 2

Given equations are x - y= 2  ...(i)
and 3x+ 2y = 16  ...(ii)
On substituting y=x- 2from Eq. (i) in Eq. (ii), we get
x = 4
y=2

\(x^{2}-(2 \alpha+3 \beta+3 \alpha+2 \beta) x+(2 \alpha+3 \beta)(3 \alpha+2 \beta)\) 
\(=x^{2}-(5 \alpha+5 \beta) x+\left(6 \alpha^{2}+4 \alpha \beta+9 \alpha \beta+6 \beta^{2}\right)\) 
\(=x^{2}-5(\alpha+\beta) x+\left[6\left(\alpha^{2}+\beta^{2}\right)+13 \alpha \beta\right]\) 
= x² - 15x+ 52

∵ \(\lambda x+y=\lambda ^{ 2 }\)
\(x+\lambda y=1\)
(i) For no solution,
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
⇒ \(\frac { \lambda }{ 1 } =\frac { 1 }{ \lambda } \neq \frac { { \lambda }^{ 2 } }{ 1 } \Rightarrow \frac { 1 }{ \lambda } \)
⇒ \({ \lambda }^{ 2 }=1\quad or\quad 1\neq { \lambda }^{ 3 }\)
⇒ \(\lambda =\pm 1\)
(ii) For infinitely many solutions,
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \Rightarrow \frac { \lambda }{ 1 } =\frac { 1 }{ \lambda } =\frac { { \lambda }^{ 2 } }{ 1 } \)
⇒ \({ \lambda }^{ 2 }=1\quad or\quad { \lambda }^{ 3 }=1\)
⇒ \(\lambda =\pm 1\quad or\quad \lambda =1,\omega ,{ \omega }^{ 2 }\)
where \(\omega \) and \({ \omega }^{ 2 }\) are cube roots of unity.
(iii) For a unique solution,
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ { b }_{ 2 } } \Rightarrow \frac { \lambda }{ 1 } \neq \frac { 1 }{ \lambda } \)
⇒ \({ \lambda }^{ 2 }\neq 1\)
⇒ \(\lambda \neq \pm 1\)

2y - x = 8

5y - x = 14

y - 2x = 1

The vertices of the triangle fromed are (1, 3), (-4, 2) and (2, 5)

p(y) = 6y² - 7y + 2
\(\alpha +\beta =-\left( -\frac { 7 }{ 6 } \right) =\frac { 7 }{ 6 } \)
and \(\alpha\beta=\frac{2}{6}=\frac{1}{3}\)
Now \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \alpha +\beta }{ \alpha \beta } =\frac { { 7 }/{ 6 } }{ { 2 }/{ 6 } } =\frac { 7 }{ 2 } \)
and \(\frac { 1 }{ \alpha } \times \frac { 1 }{ \beta } =\frac { 1 }{ \alpha \beta } =3\)
The required polynomial is \({ y }^{ 2 }-\frac { 7 }{ 2 } y+3=\frac { 1 }{ 2 } \left[ 2{ y }^{ 2 }-7y+6 \right] \)

Let \(f(y)=7y^{ 2 }-\frac { 11 }{ 3 } y-\frac { 2 }{ 3 }\)
 \(=\frac { 21y^{ 2 }-11y-2 }{ 3 } \)
\(=\frac { 21y^{ 2 }-14y+3y-2 }{ 3 } \)  [by splitting the middle term]
\(=\frac { 7y(3y-2)+1(3y-2) }{ 3 } \)
\(=\frac { 1 }{ 3 } (3y-2)(7y+1)\)
So, the value of \(7y^{ 2 }-\frac { 11 }{ 3 } y-\frac { 2 }{ 3 } \) is zero whe 3y-2=0 or 7y+1=0, i.e. when \(y=\frac { 2 }{ 3 } \) or \(y=\frac { 1 }{ 7 } \).
Thus, the zeroes\(=\frac { 2 }{ 3 } -\frac { 1 }{ 7 } =\frac { 14-3 }{ 21 } =\frac { 11 }{ 21 } =-\left( \frac { -11 }{ 3\times 7 } \right) \)
\(=-(1).\left( \frac { Coefficient \ of \ y }{ Coefficient \ of \ y^{ 2 } } \right) \) and product of zeroes\(=\left( \frac { 2 }{ 3 } \right) \left( -\frac { 1 }{ 7 } \right) =\frac { -2 }{ 21 } =\frac { -2 }{ 3\times 7 } \)
\(=\left( \frac { Constant \ term }{ Coefficient \ of \ y^{ 2 } } \right) \)
Hence, the relations between the zeroes and the coefficients of the polynomials is verified.

Let other zeroes be a.
Then, according to the given condition,
a+7=-18
⇒ a=-25
Required quadratic polynomial will be (x-7)(x+25)
or x²+18x-175

Since, a and β are the zeroes of polynomial x² -px-p+c.
So, sum of zeroes, a+β =p      ...(i)
and product of zeroes, aβ =c-p|   ...(ii)
Also, (a+1)(β+1)=0  [given]
aβ+(a+β)+1=0
⇒ c-p+p+1=0 [from Eqs. (i) and (ii)]
⇒ c-1

Let the quadratic polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+\beta=-\frac{b}{a}=1\) 
\(\alpha \beta=\frac{c}{a}=1\) 
If a = 1, b = -1, and c = 1
Therefore, the quadratic polynomial is x² – x +1.

For unit's digit to be 0, then 4ⁿ should have 2 and 5 as its prime factors, but 4ⁿ=(2²)ⁿ = 2²ⁿ.lt does not contain 5 as one of its prime factors.
ஃ 4ⁿ will not end with digit 0 for n \(\in \) N.

\({ x }^{ 2 }-4\sqrt { 3 } x+3=0\)
If \(\alpha\) and \(\beta\) are the zeroes of \({ x }^{ 2 }-4\sqrt { 3 } x+3\)
then \(\alpha+\beta=-\frac{b}{a}\)
\(\Rightarrow \quad \alpha +\beta =-\frac { \left( -4\sqrt { 3 } \right) }{ 1 } \)
\(\Rightarrow \quad \alpha +\beta=4\sqrt { 3 }\)
and \(\alpha\beta=\frac{c}{a}\)
\(\Rightarrow \quad \alpha\beta=\frac{3}{1}\)
\(\Rightarrow \quad \alpha\beta=3\)
\(\therefore \quad \alpha +\beta -\alpha \beta =4\sqrt { 3 } -3\)

Since, - 1 is a zero of the polynomial
p(x) = kx² - 4x + k,
then p(-1) = 0
\(\therefore\)  k(-1)² - 4 (- 1) + k = 0
\(\Rightarrow\)  k + 4 + k = 0
\(\Rightarrow\)  2k + 4 = 0
\(\Rightarrow\)  2k = - 4
Hence, K = - 2

Here a₁ = 2, b₁ = 1, c₁ = 3
and a₂ = 4, b₂ = 2, c₂ = 6
Clearly \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
i.e.\(\frac { 2 }{ 4 } =\frac { 1 }{ 2 } =\frac { 3 }{ 6 } \)
Hence lines are coincident.

x=3, y=-3

k=-3/2

Intersect at a point

Since, x = a and y = b is the solution of the equations x - y = 2 and x + y = r, therefore these values will satisfy that equations . Thus, we have
a - b = 2  ..(i)
and a+b=4  ...(ii)
On adding Eqs. (i) and (ii), we get
2a = 6 ⇒ a = 3
On substituting a=3 in Eq. (i), we get
3 - b = 2 ⇒ b = 1
Hence, a = 3 and b = 1

Let the number of Rs.100 and Rs.50 notes be x and y, respectively. Then, according to the question, x+y=14100x+50y=1000
Rs.100 notes=6, Rs.50 notes=8

Let x and y be length and breadth of rectangle.
Then, its area=xy
According to the questions,
9x-5)(y+2)=xy-80 \(\Rightarrow\) 2x-5y=-70
(x+10)(y-5)=xy+50  \(\Rightarrow\) -5x+10y=100
Length=40 units, breadth=30 units

\(-\frac { 13 }{ 2 } \)

\(-\frac { 145 }{ 12 } \)

x²-6x+7

α+β+αβ=\(-\frac { 7 }{ 2 } +\frac { 5 }{ 2 } =-1\)

 Quadratic