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Mathematics

Find the values of k so that the quadratic equation (4 - k)x2 + 2(k + 2)x + (8k + 1) = 0 has equal roots.

Quadratic Equations

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Answer

The given equation is (4 - k)x2 + 2(k + 2)x + (8k + 1) = 0

Comparing it with ax2 + bx + c = 0
a= (4 - k), b = (2k + 4), c = (8k + 1)

Given,

Equation has real and equal roots

∴ b2 - 4ac = 0

⇒(2k+4)2−4×(4−k)×(8k+1)=0⇒(4k2+16+16k)−4(32k+4−8k2−k)=0⇒4k2+16+16k−128k−16+32k2+4k=0⇒4k2+32k2+16k−128k+4k+16−16=0⇒36k2−108k=0⇒36k(k−3)=0⇒36k=0 or k−3=0k=0 or k=3\Rightarrow (2k + 4)^2 - 4 \times (4 - k) \times (8k + 1) = 0 \\[1em] \Rightarrow (4k^2 + 16 + 16k) - 4(32k + 4 - 8k^2 - k) = 0 \\[1em] \Rightarrow 4k^2 + 16 + 16k - 128k - 16 + 32k^2 + 4k = 0 \\[1em] \Rightarrow 4k^2 + 32k^2 + 16k - 128k + 4k + 16 - 16 = 0 \\[1em] \Rightarrow 36k^2 - 108k = 0 \\[1em] \Rightarrow 36k(k - 3) = 0 \\[1em] \Rightarrow 36k = 0 \text{ or } k - 3 = 0 \\[1em] k = 0 \text{ or } k = 3

Hence , the value of k are 0, 3.

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