For the above expression to be a perfect square, D = b2 – 4ac = 0⇒ (2k + 4)2 – 4 × (4 – k)(8k + 1) = 0⇒ 4k2 + 16k + 16 + 32k2 – 124k – 16 = 0⇒ 36k2 – 108k = 0⇒ 36k(k – 3) = 0⇒ k = 0, 3

Q. 44.6( 18 Votes )

For what value of

Class 10th RD Sharma - Mathematics 8. Quadratic Equations

Answer :

For the above expression to be a perfect square, D = b² – 4ac = 0

⇒ (2k + 4)² – 4 × (4 – k)(8k + 1) = 0

⇒ 4k² + 16k + 16 + 32k² – 124k – 16 = 0

⇒ 36k² – 108k = 0

⇒ 36k(k – 3) = 0

⇒ k = 0, 3

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