Question

If a and ß are the zeroes of the polynomial x² + kx +45, such that
(a – ß)²= 144, find k. Also find the zeroes of the polynomial.​

Answer 1

Answer:

x² + kx +45=0

(a – ß)²= 144

${(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab$

${(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \: + 2ab \: - 2ab$

${(a - b)}^{2} = {(a +b)}^{2} - 4ab$

$144 = {(a +b)}^{2} - 4ab$

From the quadratic equation

$a + b = - k$

And

$ab = 45$

So,

$144 = {k}^{2} - 4 \times 45$

$144 + 180 \: = {k}^{2}$

So,

${k}^{2} = 324$

$k = 18$

So the quadratic equation is,

x² + 18x +45 = 0

The solution is,

$x = - 3 \: or \\ x = - 15$