(K-12)x square+2(K-12)X+2=0. find the value of k for which the given equation are real and equal roots
Answers
Answer:ere, a = k – 12, b = 2 (k – 12), c = 2
∴ D = b2 – 4ac = [2(k– 12)]2
– 4(k – 12) x 2
= 4 (k – 12)2 – 8 (k– 12)
Roots are equal, if D = 0
⇒ 4 (k – 12) 2 – 8 (k – 12) = 0
⇒ 4(k – 12)(k – 12 – 2) = 0
⇒ (k – 12) (k – 14) = 0
⇒ k = 12 or 14
But k = 12 does not satisfy the eqn.
k = 14 Ans.
Step-by-step explanation:
Given:
(K-12)x²+2(K-12)x+2=0
The roots are real and equal
To find:
The value of K
Solution:
We can find the value by following the steps given below-
We know that an equation has real and equal roots when the value of its discriminant is equal to 0.
Discriminant, D= (b²-4ac)
where a is the coefficient of x², b of x, and c is the constant.
So, a= K-12, b=2(K-12), c=2
On putting the values, we get the following
D=[2(K-12)]²- 4×(K-12)×2
D=4(K-12)²-8(K-12)
D=4(K²+144-24K) -8K+96
D=4K²+576-96K-8K+96
D=4K²-104K+672
Now, D=0
4K²-104K+672=0
K²-26K+168=0
K=12, 14
On putting the values of K in the equation, only K=14 satisfies it.
Therefore, the value of k for which the given equation has real and equal roots is 14.