Question

find K $(k-5)x^2 + 2(k-5) + 2 =0$ has real and equal roots.

Answer 1

Answer:

hence k is 5 or 13

HOPE IT HELPS U

Answer 2

Answer:

7

Step-by-step explanation:

Given:

$(k-5)x^2 + 2(k-5) + 2 =0$ has real and equal roots.

To find:

The value of k

Solution:

We can solve by using Discriminat .

What is discriminat ?

The discriminant is the part of the quadratic equation in which (coefficient of x)²- 4(coefficient of x²)(constant). The discriminant tells us whether there are two solutions, one solution, or no solutions.

When the quadratic equation have equal roots then the discriminant is equal to zero.

So,

Comparing the equation (k-5)²x²+2(k-5)x+2=0 from ax² + bx + c = 0, I get

a = (k-5) b=2(k-5) c = 2

Since, Equation have equal roots

Therefore,

${b}^{2} - 4ac = 0 \\ \\ \{2(k - 5) \} {}^{2} = 4.(k - 5).2 \\ \\ 4(k - 5)(k - 5) = 4.2.(k - 5) \\ \\ \cancel{4} \cancel{(k - 5)}(k - 5) = \cancel{4}.2.\cancel{(k - 5)} \\ \\ k - 5 = 2 \\ \\ k = 2 + 5 \\ \\ k = 7$

Hence, The required value of k is 7.