Question

If the roots of the equation x^2-16(k-8)^2 x+(k^2+16k+64)=0 are reciprocal to each other then k=

Answer 1

EXPLANATION.

Quadratic equation.

⇒ x² - 16(k - 8)²x + (k² + 16k + 64) = 0.

As we know that,

Let one roots be = α.

Other roots be reciprocal to other = 1/α.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ α x 1/α = (k² + 16k + 64)/1.

⇒ 1 = k² + 16k + 64.

⇒ k² + 16k + 63 = 0.

Factorizes the equation into middle term splits, we get.

⇒ k² + 9k + 7k + 63 = 0.

⇒ k(k + 9) + 7(k + 9) = 0.

⇒ (k + 7)(k + 9) = 0.

⇒ k = -7  and  k = -9.

                                                                                                                         

MORE INFORMATION.

Nature of the roots of the quadratic epression.

(1) = Real and different, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

Answer 2

Answer:

$\huge\{ \underline \red{SOLUTION} \}$

Given,

$= > {x}^{2} - 16(k - 18 {)}^{2} x + ( {k}^{2} + 16k + 64) = 0$

Now,

Let's assume the one root as

$= > \alpha$

$other \: root \: as \: \: \frac{1}{ \alpha }$

( b'coz reciprocal of the root)

Let's find the Products of the zeros of the quadratic equation.

$\alpha \beta = \frac{c}{a}$

$= > \alpha \times \frac{1}{ \alpha } = \frac{( {k}^{2} + 16k + 64)}{1}$

$= > 1 = {k}^{2} + 16k + 64$

$= > {k}^{2} + 16k + 63 = 0$

Now, by factorising the equation into middle term split , we get

$= > {k}^{2} + 9k + 7k + 63 = 0$

$= > k(k + 9) + 7(k + 9) = 0$

$(k + 7)(k + 9) = 0$

$k = - 7 \: \: \: \: \: and \: \: \: \: \: \: k = - 9$

Therefore , the value of k is

$\pink{k = - 7 \: \: , \: \: - 9}$