Question

If one zeroes of the polynomial (a^2+9)x^2+13x+6a is reciprocal of the other find a

Answer 1

Answer:

$\mathfrak{\large \red{\underline{\underline{Answer}}}} \\ \large \blue{a = 3}$

Step-by-step explanation:

$\mathfrak{\large \red{\underline{\underline{given}}}} \\ \large \green{one \: zeroes \: of \: the \: polynomial \: is \: reciprocal \: to \: other} \\ \large \green{equation \: written \: as \: \alpha = \frac{1}{ \alpha } } \\ \mathfrak{\large \red{\underline{\underline{step \: by \: step \: explaination}}}} \\ \large \green{ ({a}^{2} + 9) {x}^{2} + 13x + 6a } \\ \large \green{product \: of \: zeroes = \alpha \times \frac{1}{ \alpha } = \frac{6a}{ {a}^{2} - 9 } } \\ \large \green{1 = \frac{6a}{ {a}^{2} - 9 } } \\ \large \green{ {a}^{2} - 6a + 9 = 0 } \\ \large \green{ {a}^{2} - 3a - 3a + 9 = 0 } \\ \large \green{a(a - 3) - 3(a - 3) = 0} \\ \large \green{(a - 3)(a - 3) = 0} \\ \large \green{(a - 3) {}^{2} = 0 } \\ \large \green{a = 3}$

Answer 2

QUESTION:

If one zeroes of the polynomial (a^2+9)x^2+13x+6a is reciprocal of the other find a

ANSWER:

Let one zero of the polynomial be

$\huge\orange { \alpha}$

then the other zero of the polynomial will be

$\huge\red{\frac{1}{ \alpha } }$

now we use the formula here product of zeroes.

we know that;

quadratic polynomial is in the form of

$\huge\blue {a {x}^{2} + bx + c = 0}$

and;

$\red {product \: of \: zeroes = \frac{ constant term}{coefficient \: of \: {x}^{2} } = \frac{ c}{a} }$

now come to main question;

here;

a = a^2+9

b = 13

c = 6a

so,

$\alpha \times \frac{1}{ \alpha } = \frac{ 6a}{ {a}^{2} + 9}$

$1 = \frac{6a}{ {a}^{2} + 9 }$

$(cross \: multiplication)$

$6a = {a}^{2} + 9$

${a}^{2} - 6a + 9 = 0$

now it is a quadratic equation we have to factorised it.

splitting the middle term in such a way that it's sum equal to -6 and product equal to 9.

${a}^{2} - 3a - 3a + 9 = 0 \\ a(a - 3) - 3(a - 3) = 0 \\ (a - 3)(a - 3) = 0$

$\huge\purple {(a - 3)(a - 3) = 0}$

so,

$a - 3 = 0 \\ a = 3$

FINAL ANSWER :

$\huge\red {a = 3}$