Question:hard

\(a\), \(b\) and \(c\) are the sides of a triangle. The equations \(ax^2 + bx + c = 0\) and \(3x^2 + 4x + 5 = 0\) have a common root. Then angle C is equal to

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The equation 3x^2+4x+5=0 has no real roots, so any real-coefficient quadratic sharing one of its roots must share both, making a, b, c proportional to 3, 4, 5, a Pythagorean triple.
Updated On: Jul 13, 2026
  • \(60^{\circ}\)
  • \(90^{\circ}\)
  • \(120^{\circ}\)
  • None of these
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The Correct Option is B

Solution and Explanation

Step 1: Establish that a, b, c are proportional to 3, 4, 5.
The equation \(3x^2+4x+5=0\) has discriminant \(16 - 60 = -44 < 0\), so its two roots are complex conjugates of each other, not real numbers. A real-coefficient quadratic like \(ax^2+bx+c=0\) can only pick up one of these as a root if it also picks up the other, since its own roots must either both be real or form a conjugate pair. Sharing exactly one complex root is not possible for a real quadratic on its own, so both roots match, forcing \(ax^2+bx+c=0\) and \(3x^2+4x+5=0\) to be scalar multiples of each other:
\[ a = 3k, \quad b = 4k, \quad c = 5k \]

Step 2: Apply the Law of Cosines directly.
The Law of Cosines states:
\[ \cos C = \frac{a^2+b^2-c^2}{2ab} \]

Step 3: Substitute a = 3k, b = 4k, c = 5k.
\[ \cos C = \frac{(3k)^2+(4k)^2-(5k)^2}{2(3k)(4k)} = \frac{9k^2+16k^2-25k^2}{24k^2} \]

Step 4: Simplify the numerator.
\[ 9k^2+16k^2-25k^2 = 0 \]
So:
\[ \cos C = \frac{0}{24k^2} = 0 \]

Step 5: Solve for angle C.
\(\cos C = 0\) means \(C = 90^{\circ}\), the only angle between 0 degrees and 180 degrees whose cosine is zero.
\[ \boxed{90^{\circ}} \]
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