Question

If the sum of the zeros of the polynomial p(x) = (a²+a)x²+ 45x + 6a is reciprocal of the other, find the value of a.​

Answer 1

$\Huge\bf\maltese{\underline{\green{Answer°᭄}}}\maltese$

$\implies$$\large\bf{\underline{\red{VERIFIED✔}}}$

$Product \: of \: zeroes=ac</p><p>\alpha\times\frac{1}{\alpha} \\ \\ =\frac{6a}{a^2+9}α×α1 \\ \\ =a2+96a</p><p>a^2+9 \\ \\ =6aa2+9 \\ \\ =6a</p><p>a^2-6a+9 \\ \\ =0a2−6a+9 \\ \\ =0</p><p>a^2-3a-3a+9 \\ \\ =0a2−3a−3a+9 \\ \\ =0a(a-3)-3(a-3) \\ \\ =0a(a−3)−3(a−3) \\ \\ =0(a-3)(a-3) \\ \\ =0(a−3)(a−3) \\ \\ =0 \\ \\ ⇒ a - 3 = 0 \\ \\ ⇒ a = 3</p><p>Therefore, Value of a is 3.</p><p>$

$\boxed{I \:Hope\: it's \:Helpful}$

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Answer 2

Answer:

Productofzeroes=ac</p><p>α×

α

1

=

a

2

+9

6a

α×α1

=a2+96a</p><p>a

2

+9

=6aa2+9

=6a</p><p>a

2

−6a+9

=0a2−6a+9

=0</p><p>a

2

−3a−3a+9

=0a2−3a−3a+9

=0a(a−3)−3(a−3)

=0a(a−3)−3(a−3)

=0(a−3)(a−3)

=0(a−3)(a−3)

=0

⇒a−3=0

⇒a=3</p><p>Therefore, Valueofais3.</p><p>