1.

For what values of K will the quadratic (4x+1)x2+(k+1)x+1=0 have equal roots.

Answer»

Given quadratic equation:

(k+4)x2+(k+1)x+1=0.

Since the given quadratic equation has equal roots, its discriminant should be zero.

∴ D = 0

⇒ (k+1)2−4 × (k+4) × 1=0

⇒ k2+2k+1−4k−16=0

⇒ k2−2k−15=0

⇒k2−5k+3k−15=0

⇒(k−5)(k+3)=0

⇒k−5=0 or k+3=0

⇒k=5 or −3

Thus, the values of k are 5 and −3.

For k = 5:

(k+4)x2+(k+1) x+1=0

⇒ 9x2+6x+1=0

⇒ (3x)2+2(3x)+1=0

⇒ (3x+1)2=0

⇒ x=−1/3, −1/3

For k = −3:

(k+4)x2+(k+1)x+1=0

⇒ x2−2x+1=0

⇒ (x−1)2=0

⇒ x=1,1

Thus, the equal root of the given quadratic equation is either 1 or −13.



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