45.0k views
2 votes
Kx²-4x(k+1)+8k=7 has two equal roots, find the value of k ​

User WPInfo
by
5.7k points

2 Answers

3 votes

Answer:

k= 7+4x/8raised to the power of 2−4x+8

Explanation:

i dont know how to be more clearer

User Ebynum
by
6.6k points
5 votes

Answer:

k = 4.

Explanation:

For equal roots the discriminant b^2 - 4ac = 0, so for

kx² - 4x(k+1) + 8k - 7 = 0

b^2 - 4ac = (-4(k+1)^2 - 4 * k ( 8k - 7) = 0

16(k^2 + 2k + 1) - 32k^2 + 28k = 0

- 16k^2 + 60k + 16 = 0

-4k^2 + 15k + 4 = 0

4k^2 - 15k - 4 = 0

(4k + 1)(k - 4) = 0

So k = -0.25 or 4.

Check by substituting k = 4 into the original equation:

4x^2 - 4*(5)x + 32 - 7 = 0

4x^2 - 20x + 25 = 0

(2x - 5)^2 = 0 .

This has equal roots so k = 4 is a correct value.

For k = -0.25:

-0.25x^2 -4*-0.25(-0.25+1)x + 8*-0.25 - 7

-0.25x^2 - 1.25x - 9 = 0

This has complex roots:

b^2 - 4ac = (-1.25)^2 - 4 * -0.25 * -9

= -7.44 so this hasn't got equal roots.

Answer is k = 4.

User Daniel Brockman
by
5.5k points