Question

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Answer 1

Given

α and β are the zeroes of the polynomial 2x²+5x-9

✏To Find:

• Value of α²-β²

☯Solution☯

Let x be the zero of given polynomial, then

$\longrightarrow\sf{p(x)= 2x^{2} +5x-9=0}$

On taking 2 common from the polynomial, we get

$\longrightarrow\sf{2\bigg(x^{2} +\dfrac{5}{2}x-\dfrac{9}{2}\bigg) =0}$

$\longrightarrow\sf{x^{2} -\Big(-\dfrac{5}{2}\Big)x+\Big(-\dfrac{9}{2}\Big)=0}$

Now, we know that general quadratic equation is expressed as

x² -(Sum of Zeroes)x+(Product of Zeroes)=0

So, on comparing our equation with general equation we get,

$\sf{Sum\:of\:Zeroes=\alpha+\beta=-\dfrac{5}{2}}$

$\sf{Product\:of\:Zeroes=\alpha\beta=-\dfrac{9}{2}}$

Also, we know that

• $\sf{a^{2} +b^{2} =(a+b)^{2}-2ab}$

• $\sf{(a^{2}-b^{2})^{2}=(a^{2}+b^{2})^{2}-(2ab)^{2}}$

So,

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

$\sf{\alpha^{2}+\beta^{2}=\bigg(-\dfrac{5}{2}\bigg)^{2}-2\bigg(-\dfrac{9}{2}\bigg)}$

$\sf{\alpha^{2}+\beta^{2}=\dfrac{25}{4}+9}$

$\sf{\alpha^{2}+\beta^{2}=\dfrac{61}{4}}$

Now,

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

$\sf{(\alpha ^{2}-\beta ^{2})^{2}=\bigg(\dfrac{61}{4}\bigg)^{2}-\bigg(2\times\dfrac{-9}{2} \bigg)^{2}}$

$\sf{(\alpha ^{2}-\beta ^{2})^{2}=\dfrac{3721}{16}-81}$

$\sf{(\alpha ^{2}-\beta ^{2})^{2}=\dfrac{2425}{16}}$

$\sf{\alpha ^{2}-\beta ^{2}=\sqrt{\dfrac{2425}{16}}}$

$\sf{\alpha ^{2}-\beta ^{2}=\pm\dfrac{\sqrt{2425}}{4}}$

I think your question requires correction because the answer does not match with given options.

Answer 2

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❚ QuEstiOn ❚

If $\alpha$ and $\beta$ are the Zeroes of the quadratic equation 2x²+5x-9 , then find the value of , $(\alpha^2-\beta^2)=?$

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❚ ANsWeR ❚

✺ Explanation :

$\implies 2x^2+5x-9=0$

$\implies \dfrac{2x^2+5x-9}{2}=\dfrac{0}{2}$

$\implies \dfrac{2x^2}{2}+\dfrac{5x}{2}-\dfrac{9}{2}=\dfrac{0}{2}$

$\implies x^2+\dfrac{5}{2}x-\dfrac{9}{2}=0$

Comparing the equation $x^2+\dfrac{5}{2}x-\dfrac{9}{2}=0$ with the equation , $x^2-(\alpha+\beta)x+\alpha\beta=0$ we get ,

✏ $(\alpha+\beta)=-\dfrac{5}{2}$

✏ $(\alpha\beta)=-\dfrac{9}{2}$

!!✺ We have to Find :-

$(\alpha^2-\beta^2)=?$

✺ Therefore :

$\implies(\alpha^2-\beta^2)$

$\implies(\alpha-\beta)(\alpha+\beta)$

$\implies\sqrt{[(\alpha+\beta)^2-4\alpha\beta]}(\alpha+\beta)$

$\implies\sqrt{[(-\dfrac{5}{2})^2-\cancel4(-\dfrac{9}{\cancel2})]}(-\dfrac{5}{2})$

$\implies\sqrt{[\dfrac{25}{4}+18]}(-\dfrac{5}{2})$

$\implies\sqrt{[\dfrac{25+72}{4}]}(-\dfrac{5}{2})$

$\implies\dfrac{-5\sqrt{97}}{2\times2}$

$\implies\dfrac{-5\sqrt{97}}{4}$

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