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प्रश्न
Show that the equation x2 + ax – 1 = 0 has real and distinct roots for all real values of x.
[Hint: D = (a2 + 4) > 0.]
योग
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उत्तर
Given: Consider the quadratic equation x2 + ax – 1 = 0
Step-wise calculation:
1. Compare with the standard form Ax2 + Bx + C = 0:
A = 1, B = a, C = –1
2. Compute the discriminant D = B2 – 4AC
= a2 – 4(1)(–1)
= a2 + 4
3. For every real a, a2 ≥ 0, so a2 + 4 ≥ 4 > 0.
Hence, D = a2 + 4 > 0 for all real a.
4. If the discriminant D > 0, a quadratic has two real and distinct roots.
Since D = a2 + 4 > 0 for every real a, the equation x2 + ax – 1 = 0 has two real and distinct roots for all real values of a.
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