Question

If the squared difference of the zeroes of the quadratic polynomial x2+kx+30 is equal to 169 find the value of k and the zeroes

Answer 1

ANSWER

$x = 2 \: \: or \: \: x = 15$

Or

$x = - 2 \: \: or \: \: x = - 15$

EXPLANATION

The given polynomial is

$f(x) = {x}^{2} + kx + 30$

where a=1,b=k, c=30

Let the zeroes of this polynomial be m and n.

Then the sum of roots is

$m + n = - \frac{b}{a} = -k$

and the product of roots is

$mn = \frac{c}{a} = 30$

The square difference of the zeroes is given by the expression.

$( {m - n})^{2} = {(m + n)}^{2} - 4mn$

From the question, this difference is 169.

This implies that:

$( { - k)}^{2} - 4(30) = 169$

${ k}^{2} -120= 169$

$k^{2} = 289$

$k= \pm \sqrt{289}$

$k= \pm17$

We substitute the values of k into the equation and solve for x.

$f(x) = {x}^{2} \pm17x + 30$

$f(x) = (x \pm2)(x \pm 15)$

The zeroes are given by;

$(x \pm2)(x \pm 15) = 0$

$x = \pm2 \: \: or \: \: x = \pm 15$