Question

Find the values of k for which quadric equation x2+2√2kx+18=0 has equal root

Answer 1

Answer:

3 or -3

Step-by-step explanation:

Given the quadratic equation x²+(2√2)kx+18 =0 ........ (1)

Now, if ax²+bx+c =0 ....... (2)  is a quadratic equation then, it's roots are given by

x= $\frac{-b+\sqrt{b^{2}-4ac } }{2a}$ and x= $\frac{-b-\sqrt{b^{2}-4ac } }{2a}$.

So, the two roots will be equal only when (b²-4ac)=0

Now, comparing equations (1) and (2) we can write, the roots of equation (1) will be equal only when,

[(2√2)k]²-4*18 =0

⇒8k² = 72

⇒ k²=9

⇒ k= 3 or -3 (Answer)

Answer 2

Answer:

Given :

p ( x ) = x² + 2 √ 2 k x + 18 = 0

For real roots :

b² - 4 a c = 0

( 2 √ 2 k )² - 4 × 1 × 18 = 0

4 × 2 × k² - 72 = 0

8 k² = 72

k² = 9

k = √ 9

k = ± 3 .

Therefore ,  the value is ± 3 for equal root of p ( x ) .