Question

Heya!

If $\alpha$ and $\beta$ in the equation are ${x}^{2} - 5x - 6 = 0$, then find the value of ${\alpha}^{2} - {\beta}^{2}$
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Answer 1

Answer:

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Answer 2

Answer:

✩ Given Polynomial : $\sf{x}^{2} - 5x - 6 = 0$

Given that $\alpha$ and $\beta$ are the Zeroes of the Polynomial.

$\rule{100}{0.8}$

☯ $\underline{\boldsymbol{Zeroes\: of \:the\: given\: Polynomial :}}$

$:\implies\sf x^2-5x-6=0\\\\\\:\implies\sf x^2-(6-1)x-6=0\\\\\\:\implies\sf x^2-6x+x-6=0\\\\\\:\implies\sf x(x-6)+1(x-6)=0\\\\\\:\implies\sf (x - 6)(x + 1)=0\\\\\\:\implies\sf \alpha = 6\:or\:-\:1 \qquad\beta = -\:1\:or\:6$

$\rule{150}{1.5}$

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$\bigstar\sf\:when\:\:\alpha=6\:and\:\beta=-\:1\\\\\dashrightarrow\sf\:\:{\alpha}^{2} - {\beta}^{2}\\\\\\\dashrightarrow\sf\:\:(6)^2-(-\:1)^2\\\\\\\dashrightarrow\sf\:\:36 - 1\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf 35}}\\\\\\\bigstar\sf\:when\:\:\alpha=-\:1\:and\:\beta=6\\\\\dashrightarrow\sf\:\:{\alpha}^{2} - {\beta}^{2}\\\\\\\dashrightarrow\sf\:\:(-\:1)^2-(6)^2\\\\\\\dashrightarrow\sf\:\:1 - 36\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf -\:35}}$

⠀

$\therefore\:\underline{\textsf{Hence, required value of the given is \textbf{ \pm 35}}}.$