Question

Find the value(s) of ݇ for which the quadratic equation ݔ
x^2 + 2√2kx +18 =0 has equal
roots.

Answer 1

Answer:

Value of k = ±3

Step-by-step explanation:

Compare x²+2√2kx+18=0 with ax²+bx+c=0, we get

a = 1 , b =2√2k, c= 18,

Discreminant (D)=0

/* Equal root */

=> b² - 4ac = 0

=> (2√2k)²-4×1×18=0

=> 8k² - 72=0

=> 8(k²-9)=0

=> k²-9 = 0

=> k² = 9

=> k = ±√9

=> k = ±3

Therefore,

Value of k = ±3

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Answer 2

Answer:

For k = ±3, the given equation has equal roots.

Step-by-step explanation:

Given quadratic equation-

$x^{2} + 2\sqrt{2}kx + 18 = 0$

Comparing with $ax^{2} +bx+c = 0$

a = 1, $b = 2\sqrt{2} k$ and c = 18

We know that-

A quadratic equation has equal roots if-

$b^{2} - 4ac = 0$

So, $b^{2} = 4ac$

Put the values of a, b and c

$(2\sqrt{2} k)^{2} = 4 \times 1 \times18$

$8k^{2}= 72$

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

$k^{2} = 9$

$k = \sqrt{9}$

k = ±3

Hence, for k = ±3, the given equation has equal roots.