Question

Find the value(s) of ݇ for which the quadratic equation x²+2√2kx+18 has equal roots
plz

Answer 1

On comparing the equation

x² + 2√2kx + 18 = 0

with

ax² + bx + c = 0

we get;

a = 1

b = 2√2k

c = 18

Since the equation has real and equal roots

b² - 4ac = 0

(2√2k)² - 4(1)(18) = 0

8k² - 72 = 0

${k}^{2} = \frac{72}{8}$

k² = 9

k = ±3