Question

For which values of k will the roots of 6x^2 + 6 = 4kx be real and equal

Answer 1

Step-by-step explanation:

$6 {x}^{2} + 6 = 4kx$

$6 {x}^{2} - 4kx + 6 = 0$

condition for real and equal roots is

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

here,

a = 6, b = -4k, c = 6

$( - 4k) {}^{2} - 4(6)(6) = 0$

$16 {k}^{2} - 144 = 0$

$16 {k}^{2} = 144$

${k}^{2} = \frac{144}{16}$

${k}^{2} = 9$

$k = \sqrt{9}$

k = 3

Answer 2

Answer:

For the roots to be real and equal, the condition is:

b² - 4 a c = 0

Given:

We have the equation:

6 x² + 6 = 4 k x.

Find:

We need to find the value of k.

Solution

For the roots to be real and equal, the condition is:

b² - 4 a c = 0

We have the equation:

6 x² + 6 = 4 k x

6 x²- 4 k x + 6

Now we can see that:

a = 6, b = -4 k , c = 6

So, we get that:

b² - 4 a c = 0

(-4 k)² - 4 × 6 × 6 = 0

16 k² - 144 = 0

16 k² = 144

k² = 144 / 16

k²= 9

k = ±3

Therefore , the values of k in the equation 6 x² + 6 = 4 k x for real and equal roots are ± 3 .

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