Question

For what value of k,(4-K)X²+(2K+4)X+(8K+1)=0 , is a perfect square.

Answer 1

k = 3

Step-by-step explanation:

(4-K)X²+(2K+4)X+(8K+1)=0 is a perfect square when discriminant is zero.

d = b² - 4ac

a = 4-k

b = 2k + 4

c = 8k + 1

Value of K for which d is zero will be:

(2k +4)² - 4(4-k)(8k+1) = 0

4k² + 16 + 16k -4(32k + 4 - 8k² -k) = 0

4k² + 16 + 16k -128k - 16 + 32k² +4k = 0

36k² -108k + 0 = 0

36k² = 108k

36k = 108

Therefore k = 108/36 = 3

Answer 2

Given:-

• The value of k, (4 - K)x² + (2K +4)x + (8K + 1) = 0 , is a perfect square.

To find:-

• Find the value of k...?

Solutions:-

• The given quadratic equation is (4 - K)x² + (2K +4)x + (8K + 1) = 0, and roots are root and equal.

• a = (4 - K)
• b = (2K +4)
• c = (8K + 1)

we know that

• D = b² - 4ac

Now,

The putting the value of a = (4 - K), b = (2K +4) and c = (8K + 1)

D = b² - 4ac

=> (2K +4)² - 4 × (4 - K) (8K + 1)

=> 4k² + 16k + 16 - 4(4 + 3k - 8k²)

=> 4k² + 16k + 16 - 12k + 32k²

=> 36k² - 108k + 0

=> 36k² - 108k

The given equation will have real and equal roots.

• D = 0

Thus,

=> 36k² - 108k

=> 18k(2k - 6) = 0

=> k(2k - 6) = 0

Now factroizing of the above equation

=> k(2k - 6) = 0

So, either

=> k = 0

Or

=> k = (2k - 6) = 0

=> 2k = 6

=> k = 6/2

=> 3

Hence, the value of k is 0 and 3.