Question

Find the polynomial whose zeroes are twice the zeroes of the polynomial . 2x^2+12root2x+35

Answer 1

$\sf x^2 + 12 \sqrt{2}x + 70$

Given:-

• $\sf 2x^2 + 12 \sqrt{2}x + 35$

To find:-

• The polynomial whose zeroes are twice the zeroes of the polynomial.

Solution:-

By middle term splitting method:-

★ $\sf 2x^2 + 12 \sqrt{2}x + 35$

$:\implies\sf 2x^2 + 7 \sqrt{2}x +5 \sqrt{2}x + 35$

$:\implies\sf \sqrt{2}x( \sqrt{2}x + 7) +5( \sqrt{2}x + 7)$

$:\implies\sf (\sqrt{2}x + 5)( \sqrt{2}x + 7)$

★ Zeroes of polynomial = $\sf \dfrac{-5}{ \sqrt{2}}$ and $\sf \dfrac{-7}{ \sqrt{2}}$

$\rule{200}3$

We have to find, a polynomial whose zeroes are twice the zeroes of the polynomial i.e $\sf 2 \times \bigg( \dfrac{-5}{ \sqrt{2}} \bigg)$ and $\sf 2 \times \bigg( \dfrac{-7}{ \sqrt{2}} \bigg)$ are zeroes of that unknown polynomial.

★ Sum of zeroes = $\sf 2 \times \bigg( \dfrac{-5}{ \sqrt{2}} \bigg) + 2 \times \bigg( \dfrac{-7}{ \sqrt{2}} \bigg) = -2 \bigg( \dfrac{5 + 7}{ \sqrt{2}} \bigg) = -12 \sqrt{2}$

★ Product of zeroes =$\sf 2 \times \bigg( \dfrac{-5}{ \sqrt{2}} \bigg) + 2 \times \bigg( \dfrac{-7}{ \sqrt{2}} \bigg) = 4 \bigg( \dfrac{ -5}{ \sqrt{2}} \bigg) \bigg( \dfrac{-7}{ \sqrt{2}} \bigg) = 70$

Therefore,

$:\implies$ x² - (sum of zeroes)x + (product of zeroes)

$:\implies$ x² - ( -12 $\sqrt{2}$)x + 70

$:\implies$ x² + 12$\sqrt{2}$x + 70

$\rule{200}3$