Question

find the quadratic polynomial whose zeros are double of the zeros of the polynomial 3x^2+2x-5​

Answer 1

Given to find the Quadratic whose whose zeros are the double of 3x² + 2x - 5

SoLuTiOn :-

$\sf 3x {}^{2} + 2x - 5$

Finding zeros of the polynomial by splitting the midlle term

Here product of two numbers should be -15x² and sum of two numbers should be 2x

So, split the middle term as

5x - 3x sum = 2x product = -15x²

$\sf3x {}^{2} + 2x - 5$

$\sf3x {}^{2} + 5x - 3x - 5$

$\sf3x {}^{2} - 3x + 5x - 5$

$\sf3x(x - 1) + 5(x - 1)$

$\sf(x - 1)(3x + 5)$

Equating two zero

$\sf x - 1 = 0$

$\sf x = 1$

$\sf 3x + 5 = 0$

$\sf x = \sf\dfrac{ - 5}{3}$

So, the zeros are 1 , -5/3

Since,

$\sf\alpha = 1$

$\sf\beta = \sf\dfrac{ - 5}{3}$

Double the zeros means

$\sf\alpha = 1(2)$

$\sf\alpha = 2$

$\sf\beta = \sf\dfrac{ - 5}{3} \times 2$

$\sf\beta = \sf\dfrac{ - 10}{3}$

So, Required Quadratic polynomial is

$\sf x {}^{2} - ( \sf\alpha + \sf\beta )x + \sf \alpha \sf \beta$

$\sf x {}^{2} - \bigg(2 - \dfrac{10}{3} \bigg)x + 2\times \dfrac{ - 10}{3}$

$x {}^{2} - \bigg(\dfrac{ - 4}{3} \bigg)x - \dfrac{20}{3}$

$x {}^{2} + \dfrac{4x}{3} - \dfrac{20}{3}$

Taking L.C.M

$\sf3x {}^{2} + 4x - 20 \div 3 = 0$

$\sf3x {}^{2} + 4x - 20$

Required Quadratic polynomial