Question

if alpha and beeta are the zeros of the plynomial3x square+5x-2, then from a quadrtic polynomial whose zeros are 2alpha and 2 beta​

Answer 1

Step-by-step explanation:

The required polynomial is

q(x) = x^2 + \dfrac{10}{3}x-\dfrac{8}{3}q(x)=x2+310x−38

Step-by-step explanation:

We are given the following in the question:

p(x) = 3x^2+5x-2p(x)=3x2+5x−2

To find the zeroes of the polynomial:

$$\begin{lgathered}p(x) = 3x^2+5x-2 = 0\\3x^2 + 6x-x-2 = 0\\3x(x+2)-1(x+2) = 0\\(3x-1)(x+2) = 0\\x = -2, x = \dfrac{1}{3}\\\alpha = -2\\\beta = \dfrac{1}{3}\end{lgathered}$$

New roots of polynomial are:

$$\begin{lgathered}\alpha' = 2\alpha = -4\\\beta' = 2\beta = \dfrac{2}{3}\end{lgathered}$$

Sum of roots:

$$\alpha' + \beta' = -4 + \dfrac{2}{3} = -\dfrac{10}{3}$$

Product of roots:

$$\alpha'\beta' = -4\times \dfrac{2}{3} = -\dfrac{8}{3}$$

New Polynomial:

$$\begin{lgathered}q(x) = x^2 - (\alpha' + \beta')x + \apha'\beta'\\\\q(x) = x^2 + \dfrac{10}{3}x-\dfrac{8}{3}\end{lgathered}$$

is the required polynomial.

#LearnMore

If alpha,beta are the zeroes of polynamial 6x^2-5x+7, then find the polynomial whose zeroes are 2alpha+3beta, 3alpha+2beta

Answer 2

hope it helps

plzz mark as brainliest.........

;)