Question

Two of the zeros of the polynomial
$({a}^{2} + 9) {x}^{2} + 13x - 6a$
are
$\alpha \: \:and \: \: \beta$
If
$\alpha = \frac{1}{ \beta }$
Find 'a'​

Answer 1

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Answer 2

$\huge{\mathsf{\underline{\underline{Answer:-}}}}$

$\large{\boxed{\boxed{\:a= - 3}}}$

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

Two of the zeros of the polynomial

$({a}^{2} + 9) {x}^{2} + 13x - 6a$

are $\alpha \: \:and \: \: \beta$

and $\alpha = \frac{1}{ \beta }$

$\sf{\underline{\star{To\:find:}}}$

value of 'a'.

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

Let p(x)=$({a}^{2} + 9) {x}^{2} + 13x - 6a$

Also,

$\alpha = \frac{1}{ \beta }$.......(i)

$\large\sf{Product\:of\:zeroes = \frac{constant\:term}{coefficient\:of\:{x}^{2}} }$

$\large\sf{\implies\:\alpha \times \beta = \frac{-6a}{ ({a}^{2} + 9) }}$

$\large\sf{ \implies \frac{1}{ \beta } \times \beta = \frac{ - 6a}{ ({a}^{2} + 9)}........using \:(i) }$

$\large\sf{\implies 1 = \frac{ - 6a}{ ({a}^{2} + 9)} }$

$\large\sf{ \implies {a}^{2} + 9 = - 6a }$

$\large\sf{\implies {a}^{2}+ 6a + 9 = 0 }$

$\large\sf{ \implies {a}^{2} + 2 \times \ a \times 3 +{3}^{2} = 0}$

$\large\sf{ \implies {(a +3) }^{2}= 0 }$

$\large\sf{\implies a + 3 = 0}$

$\large\sf{\implies \:a = -3}$

Hence, the value of 'a' is -3.