If the roots of the quadratic equation kx^{2} – 20x + (21 + k) = 0 are equal, then positive value of k?

- 3
- -4
- 4
- 6
- -6

Option 3 : 4

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RBI Grade B 2020: Full Mock Test

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200 Questions
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120 Mins

**Given:**

kx^{2} – 20x + (21 + k) = 0

**Concept Used:**

If a quadratic equation (ax^{2 }+ bx + c = 0) has equal roots, then discriminant should be zero i.e. b^{2} – 4ac = 0

**Calculation:**

kx^{2} – 20x + (21 + k) = 0

Therefore, b^{2} – 4ac = 0

⇒ (20)^{2} – 4(k)(21 + k) = 0

⇒ 400 – 4k(21 + k) = 0

⇒ 400 – 84k – 4k^{2} = 0

⇒ 4k^{2} + 84k – 400 = 0

⇒ k^{2} + 21k – 100 = 0

⇒ k^{2} + 25k – 4k – 100 = 0

⇒ k(k + 25) – 4(k + 25) = 0

⇒ (k – 4)(k + 25) = 0

⇒ k = -25, 4

Therefore, k = 4
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