Find the value of k for which the quadratic equation (k-12)x square +2(k-12)x+2=0 has two real equal roots
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one k equal 12
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★ QUADRATIC EQUATIONS ★
We have, (k -12)x² + 2(k -12)x + 2 = 0
Comparing this with ax² + bx + c = 0,
We have a = k - 12, b = 2(k – 12) and c = 2
We know that,∆ = b² – 4ac = [2(k - 12)]² – 4(k - 12)(2) = (2k - 24)² – 8(k – 12) = 4k² – 96k + 576 – 8k + 96 = 4k² – 104k + 672
∴ The roots of given equation are real and equal.
Now, ∆ = 0.
We have, (k -12)x² + 2(k -12)x + 2 = 0
Comparing this with ax² + bx + c = 0,
We have a = k - 12, b = 2(k – 12) and c = 2
We know that,∆ = b² – 4ac = [2(k - 12)]² – 4(k - 12)(2) = (2k - 24)² – 8(k – 12) = 4k² – 96k + 576 – 8k + 96 = 4k² – 104k + 672
∴ The roots of given equation are real and equal.
Now, ∆ = 0.
So, 4k² - 104k + 672 = 0
⇒4(k² - 26k + 168) = 0
⇒k² - 14k - 12k + 168 = 0
⇒k (k - 14) - 12 (k - 14) = 0
⇒(k - 14) (k - 12) = 0
∵ k - 14 = 0 or k - 12 = 0
Hence, k = 14 or k = 12
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