Question

Find a quadratic polynomial the sum and product of whose zeroes are √2 and -3/2 respectively.Also,find its zeroes.

Answer 1

Answer:

Step-by-step explanation:

Hope thse helps u

Answer 2

$\boxed{\huge{\bf{\red{Answer}}}}$

$Required \ Polynomial=2x^2-2\sqrt{2}x-3$

$Zeroes \ are \ (\frac{3\sqrt{2}}{2}) \ and \ (-\frac{\sqrt{2}}{2})$

Step-by-step explanation:

$Given \ sum \ and \ product \ of \ zeroes\\\\\\\alpha+\beta=\sqrt{2} \ and \ \alpha \beta=\frac{-3}{2}\\\\\\we \ know \ for \ Required \ Polynomial=x^{2}-(\alpha+\beta)x+\alpha \beta\\\\\\putting \ values \ here\\\\\\x^{2}-(\alpha+\beta)x+\alpha \beta\\\\\\x^2-\sqrt{2}x-\frac{3}{2}\\\\\\Now \ multiplying \ by \ 2 \ we \ get\\\\\\2x^2-2\sqrt{2}x-3\\\\\\So, \ Required \ Polynomial=2x^2-2\sqrt{2}x-3$$Now \ we \ have \ p(x)=2x^2-2\sqrt{2}x-3\\\\\\we \ have \ to \ find \ its \ zeroes\\\\\\p(x)=2x^2-2\sqrt{2}x-3\\\\\\By \ splitting \ mid \ term\\\\\\p(x)=2x^2-3\sqrt{2}x+\sqrt{2}x-3\\\\\\p(x)=\sqrt{2}x(\sqrt{2}x-3)+(\sqrt{2}x-3)\\\\\\p(x)=(\sqrt{2}x-3)(\sqrt{2}x+1)\\\\\\x=\frac{3}{\sqrt{2}} \ or \ x=\frac{-1}{\sqrt{2}}\\\\\\Rationalize \ the \ denominator \ we \ get\\\\\\x=\frac{3\sqrt{2}}{2} \ or \ x=-\frac{\sqrt{2}}{2}\\\\\\Thus \ we \ get \ our \ zeroes \ also.$