Question

If alpha and beta are the zeros of the polynomial p(x)=9x square + 9x + 2, find the value of alpha square + beta square.

Answer 1

$\sf\large\underline{Given:}$

• $\rm{Polynomial\:P(x)=9x^2+9x+2}$

$\sf\large\underline{To\: Find:}$

• $\rm{The\: value\:of\alpha^2+\beta^2}$

$\sf\large\underline{Solution:}$

• At first calculate the zeroes by using the formula]

$\sf\large\underline{Formula\:used:}$

$\rm{\implies \alpha+\beta=-\dfrac{b}{a}}$

• Here, a=9 , b=9 , c=2

$\rm{\implies \alpha+\beta=-\dfrac{9}{9}}$

$\rm{\implies \alpha+\beta=-1}$

$\rm{\implies \alpha\times\beta=\dfrac{c}{a}}$

$\rm{\implies \alpha\times\beta=\dfrac{2}{9}}$

• Now calculate the the value of given expression]

$\rm{\implies Here,\: using\:identies}$

$\rm{\implies a^2+b^2=(a+b)^2-2ab}$

• If we compared with above expression the condition is same in both ,so]

$\rm{\implies \alpha^2+\beta^2= (\alpha+\beta)^2-2\alpha\:\beta}$

• Now, putting the value here]

$\rm{\implies \alpha^2+\beta^2=(-1)^2-2(\dfrac{2}{9}}$

$\rm{\implies \alpha^2+\beta^2=1-\dfrac{4}{9}}$

$\rm{\implies \alpha^2+\beta^2=\dfrac{9-4}{9}}$

$\rm{\implies \alpha^2+\beta^2=\dfrac{5}{9}}$

$\sf\large{Hence,}$

$\rm{\implies The\: value\:of\alpha^2+\beta^2=\dfrac{5}{9}}$

Answer 2

Answer:

ans in in above pick

Step-by-step explanation:

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