Quantitative Aptitude

Algebra

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All Polynomial Roots and Coefficients 2451 Zero-Product Root Finding 667 Linear System Manipulation 162 Polynomial Theorems and Reduction 85 Algebraic Identity Simplification 82 Surds and Indices 70 Modulus and Absolute Value 54 Algebraic Expression Substitution 46 Algebraic Fraction Simplification 33 Logarithmic Equations 25 Repeated Series Sum 12 Linear Equation Word Problems 9

Algebra Questions

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha$ and $\beta$ are the roots of the equation $ax^{2} \, + \, bx \, + \, c \, = \, 0$. The equation whose roots are as given below.
$\alpha \, + \,\dfrac{1}{\beta} \, , \, \beta \, + \, \dfrac{1}{\alpha}$ is $acx^2 \, + \, b(a \, + \, c) \, x \, + \, (a \, + \, c)^2 \, = \, 0$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Rightarrow$  $\alpha$ and $\beta$ are roots of the equation $ax^2+bx+c=0$

$\Rightarrow$  $\alpha+\beta=\dfrac{-b}{a}$                       -------- ( 1 )
$\Rightarrow$  $\alpha\beta=\dfrac{c}{a}$                    ------- ( 2 )
Now,
$\Rightarrow$  $\alpha+\dfrac{1}{\beta}+\beta+\dfrac{1}{\alpha}=(\alpha+\beta)+\left(\dfrac{1}{\beta}+\dfrac{1}{\alpha}\right)$

                                    $=(\alpha+\beta)+\left(\dfrac{\alpha+\beta}{\alpha\beta}\right)$
                 
                                    $=\dfrac{-b}{a}+\dfrac{\dfrac{-b}{a}}{\dfrac{c}{a}}$     [ By using ( 1 ) and ( 2 ) ]

                                    $=\dfrac{-b}{a}-\dfrac{b}{c}$

                                    $=\dfrac{-bc-ba}{ac}$

$\therefore$   $\alpha+\dfrac{1}{\beta}+\beta+\dfrac{1}{\alpha}=\dfrac{-b(a+c)}{ac}$                    ----- ( 3 )

$\Rightarrow$  $\left(\alpha+\dfrac{1}{\beta}\right)\left(\beta+\dfrac{1}{\alpha}\right)=\alpha\beta+1+1+\dfrac{1}{\alpha\beta}$
 
                                            $=\dfrac{c}{a}+2+\dfrac{1}{\dfrac{c}{a}}$

                                            $=\dfrac{c}{a}+2+\dfrac{a}{c}$
 
                                            $=\dfrac{a^2+2ac+c^2}{ac}$

$\therefore$  $\left(\alpha+\dfrac{1}{\beta}\right)\left(\beta+\dfrac{1}{\alpha}\right)=\dfrac{a^2+2ac+c^2}{ac}$              ----- ( 4 )
Now, new equation,

$\Rightarrow$  $x^2-\left(\alpha+\dfrac{1}{\beta}+\beta+\dfrac{1}{\alpha}\right)x+\left[\left(\alpha+\dfrac{1}{\beta}\right)\left(\beta+\dfrac{1}{\alpha}\right)\right]=0$
By using ( 3 ) and ( 4 ),

$\Rightarrow$  $x^2+\dfrac{b(a+c)}{ac}x+\dfrac{a^2+2ac+c^2}{ac}$

$\Rightarrow$  $acx^2+b(a+c)x+(a+c)^2=0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

A quadratic polynomial $p(x)$ with $3$ and $\dfrac{-2}{5}$  as the sum and product of zeroes, respectively is $10x^2+30x-4$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$10x^2+30x-4=0$

$\alpha+\beta=-\cfrac{30}{10}=-3$
$\alpha\beta=\cfrac{-4}{10}=\cfrac{-2}{5}$
So, a quadratic polynomial $p(x)$ with $3$ and $\cfrac{-2}{5}$ as the sum and product of zeroes, respectively is $10x^2+30x-4$ is false.

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If the roots of a quadratic equation are reciprocals of the roots of $ax^2 + bx + c = 0$, then what will be the coefficient of $c$?

  1. $x$
  2. $x^2$
  3. $-x$
  4. $x^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know to find the equation with reciprocal roots, we simply put $\dfrac {1}{x}$ in the place of $x$.
so, $a(\dfrac {1}{x})^2 + b(\dfrac {1}{x}) + c = 0$
On taking LCM we get the required equation
$a + bx + cx^2 = 0$
So the coefficient of $c$ is $x^2$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

Find the Quadratic Equation whose roots are Reciprocal of $ax^2 + bx + c = 0$.

  1. $ax^2 + bx + c = 0$
  2. $a(\dfrac {1}{x})^2 + b(\dfrac {1}{x}) + c = 0$
  3. $3ax^2 + 2bx + c = 0$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that for reciprocal roots, we only need to replace $x$ by $\dfrac {1}{x}$, in the given equation.
So the above equation becomes:
$a(\dfrac {1}{x})^2 + b(\dfrac {1}{x}) + c = 0$, which is the required answer.

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If A.M. of the roots of a quadratic equation is $8/5$ and A.M. of their reciprocals is $8/7$, then the equation is?

  1. $7x^2-16x+8=0$
  2. $3x^2-12x+7=0$
  3. $5x^2-16x+7=0$
  4. $7x^2-16x+5=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\Rightarrow$  Let $\alpha$ and $\beta$ are the roots of the equation.

According to the question,
$\Rightarrow$ $\dfrac{\alpha+\beta}{2}=\dfrac{8}{5}$

$\therefore$  $\alpha+\beta=\dfrac{16}{5}$                ----- ( 1 )

According to the equation,
$\Rightarrow$  $\dfrac{\dfrac{1}{\alpha}+\dfrac{1}{\beta}}{2}=\dfrac{8}{7}$
$\Rightarrow$  $\dfrac{\alpha+\beta}{2\alpha\beta}=\dfrac{8}{7}$
$\Rightarrow$  $7(\alpha+\beta)=16\alpha\beta$
$\Rightarrow$  $\dfrac{7\times \dfrac{16}{5}}{16}=\alpha\beta$
$\therefore$    $\alpha\beta=\dfrac{7}{5}$          - ------ ( 2 )
Now, new eqution,
$\Rightarrow$  $x^2-(\alpha+\beta)x+(\alpha.\beta)=0$
From ( 1 ) and ( 2 ),
$\Rightarrow$  $x^2-\dfrac{16}{5}x+\dfrac{7}{5}=0$
$\Rightarrow$  $5x^2-16x+7=0$   

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha, \beta$ are the root of a quadratic equation $x^2 - 3x+5=0$, then the equation whose roots are $(\alpha^2 - 3 \alpha +7)$ and $(\beta^2 -3\beta +7)$ is

  1. $x^2 +4x+1=0$
  2. $x^2 -4x+4=0$
  3. $x^2 -4x-1=0$
  4. $x^2 +2x+3=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since $\alpha, \beta$ are the root of equation $x^2-3x+5=0$
So, $\alpha^2-3\alpha +5=0$
$\beta^2 -3\beta +5=0$
$\therefore \alpha^2 -3\alpha =-5$
$\beta^2 - 3 \beta =-5$
Putting in $(\alpha^2 - 3 \alpha +7) $  &  $(\beta^2 - 3\beta +7)$    ....... (1)
$-5 +7, -5 +7$
$\therefore$ 2 and 2 are the roots
$\therefore$ The required equation is $x^2 - 4x+4=0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha+\beta$$=-2$ and ${\alpha}^{3}+{\beta}^{3}$$=-56$ then the quadratic equation whose roots are $\alpha,\beta$ is 

  1. ${ x }^{ 2 }+2x-16$$=0$
  2. ${x}^{2}+2x-15$$=0$
  3. ${x}^2+2x-12$$=0$
  4. ${x}^{2}+2x-8$$=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\Rightarrow$  $\alpha+\beta=-2$             ------ ( 1 )


$\Rightarrow$  $\alpha^3+\beta^3=-56$


$\Rightarrow$  $(\alpha+\beta)^3=\alpha^3+\beta^3+3\alpha^2\beta+3\alpha\beta^2$

$\Rightarrow$  $(\alpha+\beta)^3=\alpha^3+\beta^3+3\alpha\beta(\alpha+\beta)$

$\Rightarrow$  $(-2)^3=-56+3\alpha\beta(-2)$            [ Using ( 1 ) and ( 2 ) ]

$\Rightarrow$  $-8+56=-6\alpha\beta$

$\Rightarrow$  $48=-6\alpha\beta$

$\Rightarrow$  $\alpha\beta=-8$                      ----- ( 2 )

The required quadratic equation,

$x^2-(\alpha+\beta)x+(\alpha\beta)=0$

Using ( 1 ) and ( 3 ) we get,
$\Rightarrow$  $x^2+2x-8=0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha \neq \beta$ but $\alpha^2 = 5 \alpha -3$ and $\beta^2 = 5\beta -3$, then the equation whose roots are $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$is

  1. $3x^2 - 25x+3=0$
  2. $x^2 +5x-3=0$
  3. $x^2 -5x+3=0$
  4. $3x^2 - 19x+3=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
${\alpha}^{2}=5\alpha-3$

${\alpha}^{2}-5\alpha+3=0$

$\alpha=\dfrac{5\pm\sqrt{{5}^{2}-4\times 1\times 3}}{2}$

$\alpha=\dfrac{5\pm\sqrt{25-12}}{2}$

$\alpha=\dfrac{5\pm\sqrt{13}}{2}$

${\beta}^{2}=5\beta-3$

${\beta}^{2}-5\beta+3=0$

$\beta=\dfrac{5\pm\sqrt{{5}^{2}-4\times 1\times 3}}{2}$

$\beta=\dfrac{5\pm\sqrt{25-12}}{2}$

$\beta=\dfrac{5\pm\sqrt{13}}{2}$

Given:$\alpha\neq\,\beta$

Let $\alpha=\dfrac{5+\sqrt{13}}{2}$ and $\beta=\dfrac{5-\sqrt{13}}{2}$

$\Rightarrow\,\dfrac{\alpha}{\beta}=\dfrac{\dfrac{5+\sqrt{13}}{2}}{\dfrac{5-\sqrt{13}}{2}}$

$=\dfrac{5+\sqrt{13}}{5-\sqrt{13}}$

$=\dfrac{5+\sqrt{13}}{5-\sqrt{13}}\times \dfrac{5+\sqrt{13}}{5+\sqrt{13}}$

$=\dfrac{25+13+10\sqrt{13}}{25-13}$

$=\dfrac{38+10\sqrt{13}}{12}$

$=\dfrac{19+5\sqrt{13}}{6}$

$\Rightarrow\,\dfrac{\beta}{\alpha}=\dfrac{\dfrac{5-\sqrt{13}}{2}}{\dfrac{5+\sqrt{13}}{2}}$

$=\dfrac{5-\sqrt{13}}{5+\sqrt{13}}$

$=\dfrac{5-\sqrt{13}}{5+\sqrt{13}}\times \dfrac{5-\sqrt{13}}{5-\sqrt{13}}$

$=\dfrac{25+13-10\sqrt{13}}{25-13}$

$=\dfrac{19-10\sqrt{13}}{12}$

$=\dfrac{19-5\sqrt{13}}{6}$

Sum of the zeroes$=\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}$

$=\dfrac{19+5\sqrt{13}}{6}+\dfrac{19-5\sqrt{13}}{6}$

$=\dfrac{19+5\sqrt{13}+19-5\sqrt{13}}{6}$

$=\dfrac{2\times 19}{6}=\dfrac{19}{3}$

Product of the zeroes$=\dfrac{\alpha}{\beta}\times\dfrac{\beta}{\alpha}$

$=\dfrac{19+5\sqrt{13}}{6}\times\dfrac{19-5\sqrt{13}}{6}$

$=\dfrac{361-25\times 13}{36}=\dfrac{361-325}{36}=\dfrac{36}{36}=1$

Now,we know the foumula for finding quadratic equations,
${x}^{2}-\left(sum\,of \,the \,zeroes\right)x+product\,of\,the \,zeroes=0$

${x}^{2}-\dfrac{19}{3}x+1=0$

Hence the equation is $3{x}^{2}-19x+3=0$
Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If the difference of the roots of the quadratic equation is 3 and difference between their cubes is 189, then the quadratic equation is x2±9x+18=0x2±9x+18=0
State true or false.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the roots of the equation be a and b
then, $a^3 - b^3 = 189$
and $a - b = 3$
cubing both sides:
$(a-b)^3 = 27 $
$a^3 - b^3 - 3ab (a-b) = 27$
$189 -3ab(3) = 27 $
$162 = 9 ab$
$ab = 18$
Similarly, $(a+b)^2 = (a -b)^2  + 4ab$
$(a+b)^2 = 3^2 + 4(18)$
$(a+b)^2 = 9 + 72 $
$a +b = \pm 9$
The general form of equation is $x^2 -Sx + P = 0 $, hence the equation will be
$x^2 \pm 9x + 18 = 0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha , \beta$ are the roots of the equation $ { x }^{ 2 } - 2x + 3 = 0$, obtain the equation whose roots are ${ \alpha  }^{ 3 } - 3{ \alpha  }^{ 2 } + 5\alpha - 2,  { \beta  }^{ 3 } - { \beta  }^{ 2 } + \beta + 5$.

  1. ${x}^{2}-3x+2=0$
  2. ${x}^{2}+3x-2=0$
  3. $-{x}^{2}-3x+2=0$
  4. $-{x}^{2}+3x-2=0$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

If $\alpha, \beta$ are the roots of $x^2-2x+3=0$
then $\displaystyle \alpha ^{2}-2\alpha +3= 0$ ...(1)
and $\displaystyle \beta^2-2\beta+3=0$  ....(2)
$\displaystyle \therefore \alpha ^{2}= 2\alpha -3, \alpha ^{3}= 2\alpha ^{2}-3\alpha $
$\displaystyle \therefore P= \left ( 2\alpha ^{2}-3\alpha  \right )-3\alpha ^{2}+5\alpha -2$
$\displaystyle = -\alpha ^{2}+2\alpha -2= 3-2= 1,$ by (1)
Similarly $\displaystyle Q= 2 \therefore S= 3, P= 2$
Hence reqd. eq. is $\displaystyle x^{2}-3x+2= 0.$ or $-x^2+3x-2=0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If the difference of the roots of a quadratic equation is 4 and the difference of their cubes is 208, then the quadratic equation is $x^{2}\, \pm\, 8x\, +\, 12\, =\, 0$
State true or false.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the roots of the equation be a and b
then, $a^3 - b^3 = 208$
and $a - b = 4$
cubing both sides:
$(a-b)^3 = 64 $
$a^3 - b^3 - 3ab (a-b) = 64$
$208 -3ab(4) = 64 $
$144 = 12 ab$
$ab = 12$
Similarly, $(a+b)^2 = (a -b)^2  + 4ab$
$(a+b)^2 = 4^2 + 4(12)$
$(a+b)^2 = 16 + 48 $
$a +b = \pm 8$
The general form of equation is $x^2 -Sx + P = 0 $, hence the equation will be
$x^2 \pm 8x + 12 = 0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha$ and $\beta$ are the roots of the equation $4x^{2}\, -\, 5x\, +\, 2\, =\, 0$, find the equation whose roots are
$\alpha\, +\, \displaystyle \frac{1}{\alpha}$ and $\beta\, +\, \displaystyle \frac{1}{\beta}$.

  1. $8x^{2}\, +\, 30x\, +\, 29\, =\, 0$
  2. $x^{2}\, -\, 30x\, +\, 29\, =\, 0$
  3. $8x^{2}\, -\, 30x\, +\, 29\, =\, 0$
  4. $x^{2}\, +\, 30x\, +\, 29\, =\, 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation is: $4x^2 - 5x + 2 = 0 $


Sum of the roots = $\dfrac{5}{4}$

Product of the roots  = $\dfrac{2}{4} = \dfrac{1}{2}$

If the roots are $\alpha + \dfrac{1}{\alpha}, \beta + \dfrac{1}{\beta}$

Sum of roots = $\alpha + \dfrac{1}{\alpha} + \beta + \dfrac{1}{\beta}$
 
= $\alpha + \beta + \dfrac{\alpha+ \beta}{\alpha\beta}$

= $\dfrac{5}{4} + \dfrac{\dfrac{5}{4}}{\dfrac{1}{2}}$

= $\dfrac{5}{4} + \dfrac{5}{2}$

$\dfrac{15}{4}$

Product of roots = $(\alpha + \dfrac{1}{\alpha})( \beta + \dfrac{1}{\beta})$

= $\alpha\beta + \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} + \dfrac{1}{\alpha\beta}$

= $\dfrac{1}{2} + \dfrac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} + 2$

= $\dfrac{1}{2} + \dfrac{\dfrac{25}{16} - 1}{\dfrac{1}{2}} + 2$

= $\dfrac{1}{2} + \dfrac{9}{8} + 2$

= $\dfrac{4 + 9 + 16}{8}$ 

= $\dfrac{29}{8}$

Hence.the equation in the standard form, $x^2 - Sx + P  = 0$ can be written as:

=$x^2 - \dfrac{15}{4}x + \dfrac{29}{8} = 0$

= $8x^2 - 30x + 29 = 0$

Hence option $'C'$ is the answer.

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

Let $\alpha$ and $\beta$ be the roots of the equation ${ x }^{ 2 }+x+1=0$. The equation whose roots are ${ \alpha  }^{ 19 },{ \beta  }^{ 7 }$ is

  1. ${ x }^{ 2 }-x-1=0$
  2. ${ x }^{ 2 }-x+1=0$
  3. ${ x }^{ 2 }+x-1=0$
  4. ${ x }^{ 2 }+x+1=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$ { x }^{ 2 }+x+1=0$


$\Rightarrow \left( x-\omega  \right) \left( x-{ \omega  }^{ 2 } \right) =0$

$\Rightarrow x=\omega ,{ \omega  }^{ 2 }$

$\therefore \alpha =\omega ,\beta ={ \omega  }^{ 2 }$   $(\because \omega ,{ \omega  }^{ 2 }$ are cube roots of unity $)$

Hence, ${ \alpha  }^{ 3 }=\omega ^3 =1$
             ${ \beta  }^{ 3 }=[{\omega ^3}]^2 = 1$
             $\alpha \beta =\omega^3=1$

$\therefore { \alpha  }^{ 19 }={ \left( { \alpha  }^{ 3 } \right)  }^{ 6 }\alpha ={ 1 }^{ 6 }\alpha =\alpha =\omega $ and ${ \beta  }^{ 7 }={ \beta  }^{ 6 }.\beta ={ 1 }^{ 2 }.\beta =\beta ={ \omega  }^{ 2 }$

$\\ \Rightarrow { \alpha  }^{ 19 }+{ \beta  }^{ 7 }=\omega +{ \omega  }^{ 2 }=-1$ 
$\Rightarrow { \alpha  }^{ 19 }{ \beta  }^{ 7 }=\omega .{ \omega  }^{ 2 }={ \omega  }^{ 3 }=1$

Hence equation whose roots are ${ \alpha  }^{ 19 },{ \beta  }^{ 7 }$ is

${ x }^{ 2 }-\left( { \alpha  }^{ 19 }+{ \beta  }^{ 7 } \right) x+{ \alpha  }^{ 19 }{ \beta  }^{ 7 }=0$

$\Rightarrow { x }^{ 2 }+x+1$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

Which of the following quadratic equation has the sum of their roots $4$ and the sum of the cubes of their roots as $28$? 

  1. $x^2 - 4x + 3 = 0$
  2. $x^2 - 4x - 5 = 0$
  3. $x^2 - 3x + 4 = 0$
  4. $x^2 + 4x + 3 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $\alpha$ and $\beta$ be the roots of the equation.
Hence
$\alpha+\beta=4$
and $\alpha^{3}+\beta^{3}=28$
Now $\alpha^{3}+\beta^{3}$ can be written as

$=(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta)$
Hence
$28=64-12\alpha\beta$
$12\alpha\beta=36$
$\alpha\beta=3$
Therefore,
$x^{2}-(\alpha+\beta)x+\alpha\beta=0$
$x^{2}-(4x)+3=0$
Hence, option $A$ is correct.

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\alpha$ and $\beta$ are the roots of the equation $4x^{2}\, -\, 5x\, +\, 2\, =\, 0$, find the equation whose roots are
$\alpha\, +\, 3\beta$ and $3\alpha\, +\, \beta$.

  1. $16x^{2}\, +\, 80x\, +\, 107\, =\, 0$
  2. $16x^{2}\, -\, 80x\, +\, 107\, =\, 0$
  3. $16x^{2}\, -\, 80x\, -\, 107\, =\, 0$
  4. $16x^{2}\, +\, 80x\, -\, 107\, =\, 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$4x^2 - 5x + 2 = 0 $


If $\alpha$ and $\beta$ are the roots of this equation, 

then , sum of roots: $\alpha + \beta$ = $\displaystyle \frac{5}{4}$

Product of roots: $\alpha. \beta = \displaystyle \frac{2}{4}$

The equation which has roots as : $\alpha + 3\beta$ and $\beta + 3\alpha$

Sum of roots: $4\alpha + 4\beta$ = $4 \left (\dfrac{5}{4} \right ) = 5$

Product of roots: $(\alpha + 3\beta)(3\alpha + \beta) $

$= 3(\alpha^2 + \beta^2) + 10\alpha\beta$

$= 3(\alpha + \beta)^2 - 6\alpha\beta + 10\alpha\beta$

$= 3 \left (\dfrac{5}{4} \right )^2 + 4\frac{2}{4}$

$= \dfrac{107}{16}$

Thus new equation is :$x^2 -Sx + P = 0$


$\therefore x^2 - 5x + \dfrac{107}{16} = 0$

$\therefore 16x^2 - 80x + 107 = 0$