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For what value of k, \((4-k)\text{x}^2+(2k+4)\text{x}+(8k+1)=0,\) is a perfect square. |
Answer» \((4-k)\text{x}^2+(2k-4)\text{x}+(8k + 1)=0\) 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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