Question

α and β are zeroes of quadratic polynomial 4x² - x - 4 then the quadratic polynomial whose zeroes are 2α and 2β is :​

Answer 1

$\large\bf{\underline{\underline\red{Given:-}}}$

• α and β are zeroes of quadratic polynomial 4x² - x - 4 .

$\large\bf{\underline{\underline\blue{To\:Find:-}}}$

• The quadratic polynomial whose zeroes are 2α and 2β .

$\large\bf{\underline{\underline\green{Solution:-}}}$

Given:-

Quadratic polynomial = 4x² - x - 4 .

Comparing the polynomial with ax² + bx + c

• a = 4

• b = -1

• c = -4

$\implies \rm Sum \: of \: zeroes = \dfrac{ - b}{ \: \: a}$

$\implies \rm \alpha + \beta = \dfrac{ - ( - 1)}{ \: \: 4}$

$\implies \rm \alpha + \beta = \dfrac{ 1}{ 4}$

$\implies \rm Product \: of \: zeroes = \dfrac{c}{ a}$

$\implies \rm \alpha \beta = \dfrac{ -4}{ \: \: 4}$

$\implies \rm \alpha \beta = -1$

Now;

We need to find the quadratic polynomial of zeroes 2α and 2β

Sum of zeroes = 2α + 2β

= 2(α + β)

= $2 \times \dfrac{1}{4}$

$Sum\: of\:zeroes = \dfrac{1}{2}$

Product of zeroes= 2α × 2β

= 2(αβ)

= 2(-1)

Product of zeroes = -2

Quadratic polynomial

= x² -(sum of zeroes)x + product of zeroes

$\rm = {x}^{2} + \bigg( - \dfrac{1}{2} \bigg)x + ( - 1)$

$\rm = {x}^{2} - \dfrac{1}{2} x - 1$

$\rm = 2{x}^{2} - x - 2$

The quadratic polynomial whose zeroes are 2α and 2β is 2x² - x - 2