Question

If alpha band beta are zeros of polynomial 6x^2-7x-3,then from a quadratic polynomial where zeros are 2alpha and 2beta.

Answer 1

Step-by-step explanation:

alpha+beta = -b/a = 7/6

alpha × beta = c/a = -1/2

2(alpha + beta) = 7/3

4 alpha beta = -2

formula = x² - SUM x + PRODUCT

= x²-7/3 x -2 = 3x²-7x-6

Answer 2

Step-by-step explanation:

Given,α and β are the zeroes of the polynomial 6x² - 7x - 3

For a quadratic polynomial ax² + bx + c

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

Product of zeroes = $\dfrac{c}{a}$

In the polynomial 6x² - 7x - 3

• a = 6 , b = -7 and c = -3

$\sf \dashrightarrow \alpha + \beta = \dfrac{-b}{a} = \dfrac{-(-7)}{6} = \dfrac{7}{6}$

$\sf \dashrightarrow \alpha\beta = \dfrac{c}{a} = \dfrac{-3}{6} = \dfrac{-1}{2}$

Now, let us find the polynomial of zeroes 2α and 2β

Sum of zeroes = 2α + 2β

$\: \: \: \: \:\:\:\:\:\:\:\:\:\:\: \:\:\:\:\:\:\:\:\:\:\:\:$= 2(α + β)

$\: \: \: \: \:\:\:\:\:\:\:\:\:\:\: \:\:\:\:\:\:\:\:\:\:\:\:$ = 2 × $\dfrac{7}{6} = \dfrac{7}{3}$

Product of zeroes= 2α × 2β

$\: \: \: \: \:\:\:\:\:\:\:\:\:\:\: \:\:\:\:\:\:\:\:\:\:\:\:$= 4(αβ)

$\: \: \: \: \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \:\:\:\:\:\:$ = 4 × $\dfrac{-1}{2} = -2$

As, the general form of the quadratic equation is

➝ x² - (sum of zeroes)x + product of zeroes = 0

➝ $\sf x^{2} - \dfrac{7x}{3} - 2 = 0$

➝ $\sf \dfrac{3x^{2} - 7x - 6 }{3}= 0$

➝ $\sf 3x^{2} - 7x - 6 = 0$

∴ Required quadratic polynomial = 3x² - 7x - 6