Q. No 8Find the value of k for which the quadratic equations (k-12)x²+2(k-12)x+2=0 k²x²-2(k-1)x+4=0 have real and equal roots. About the author Jade
The given equation is (k – 12)x2 + 2(k – 12)x + 2 = 0 Here, a = k – 12, b = 2(k – 12) and c = 2 Since, the given equation has two equal real roots then we must have b2 – 4ac = 0 ⇒ [2(k – 12)]2 – 4(k – 12) x 2 = 0 ⇒ 4(k – 12)2 – 8(k – 12) = 0 ⇒ 4(k – 12) {k – 12 – 2} = 0 ⇒ (k – 12) (k – 14) = 0 ⇒ k – 12 = 0 or k – 14 = 0 ⇒ k = 12 or k = 14. Note: But at k = 12, terms of x2 and x in the equation vanish hence only k = 14 is acceptable. Reply
The given equation is
(k – 12)x2 + 2(k – 12)x + 2 = 0
Here, a = k – 12, b = 2(k – 12) and c = 2
Since, the given equation has two equal real roots
then we must have b2 – 4ac = 0
⇒ [2(k – 12)]2 – 4(k – 12) x 2 = 0
⇒ 4(k – 12)2 – 8(k – 12) = 0
⇒ 4(k – 12) {k – 12 – 2} = 0
⇒ (k – 12) (k – 14) = 0
⇒ k – 12 = 0 or k – 14 = 0
⇒ k = 12 or k = 14.
Note: But at k = 12, terms of x2 and x in the equation vanish hence only k = 14 is acceptable.