Question

It One zero of quadratic polynemial
f(x) = 4x2 - 8kx - 9 is negative of the other
find the value of k.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic polynomial is

$\rm :\longmapsto\:f(x) = {4x}^{2} - 8kx - 9$

Given that one zero of the given quadratic polynomial f(x) is negative of the other.

$\rm :\longmapsto\:Let \: zeroes \: be \: \alpha \: and \: - \alpha$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm :\implies\: \alpha + ( -\alpha ) = - \: \dfrac{( - 8k)}{4}$

$\rm :\longmapsto\:0 = 2k$

$\bf\implies \:k = 0$

Note :-

Short Cut Tricks : -

1. If one zero of the quadratic polynomial f(x) = ax² + bx + c is negative of the other, then b = 0, i.e. coefficient of x = 0.

2. If one zeroes of the quadratic polynomial f(x) = ax² + bx + c is reciprocal of the other, then a = c, i.e. coefficient of x² = constant term.

Answer 2

$\huge\bold\red{Answer}$

$f(x) = {4x}^{2} - 8kx + 8x - 9$

Let one zero be a

and then other be = - a

sum of the zero =

$= > a - a = 0 = \frac{(8k - 8)}{4}$

$= > k = 1$

Zero of

$= > 1( {x})^{2} + 3(x)x + 2$

$= > {x}^{2} + 3x + 2$

$= > {x}^{2} + x + 2x + 2$

$= > x(x + 1) + 2(x + 1)$

$= > (x + 1)(x + 2)$

Zero of

$x = - 1. - 2$

Hence, the answer is -1,-2.