Find The Values Of K For Roots Are Real And Equal In Equation: (k + 1)x^2

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Ask a Question Find the values of k for roots are real and equal in equation: (k + 1)x^2 – 2(3k + 1)x + 8k + 1 = 0 ← Prev Question Next Question → 0 votes 10.4k views asked Mar 14, 2020 in Quadratic Equations by ShasiRaj (56.4k points)

Find the values of k for roots are real and equal in equation:

(k + 1)x2 – 2(3k + 1)x + 8k + 1 = 0

• quadratic equations
• class-10

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1 Answer

0 votes answered Mar 14, 2020 by Tahseen Ahmad (30.6k points) selected Mar 14, 2020 by ShasiRaj Best answer

The given equation (k +1)x2 – 2(3k +1)x + 8k + 1 = 0 is in the form of ax2 + bx + c = 0

Where a = (k +1), b = -2(3k + 1), c = 8k + 1

For the equation to have real and equal roots, the condition is

D = b2 – 4ac = 0

⇒ (-2(3k + 1))2 – 4(k +1)(8k + 1) = 0

⇒ 4(3k +1)2 – 4(k + 1)(8k + 1) = 0

⇒ (3k + 1)2 – (k + 1)(8k + 1) = 0

⇒ 9k2 + 6k + 1 – (8k2 + 9k + 1) = 0

⇒ 9k2 + 6k + 1 – 8k2 – 9k – 1 = 0

⇒ k2 – 3k = 0

⇒ k(k – 3) = 0

Either k = 0 Or, k – 3 = 0 ⇒ k = 3,

So, the value of k can either be 0 or 3

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