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Question

Question: If \(x^{2} + x + a = 0\) are real and \(2 < a < 3\), then the roots of the equation \(a > 3\)are....

If x2+x+a=0x^{2} + x + a = 0 are real and 2<a<32 < a < 3, then the roots of the equation a>3a > 3are.

A

Complex

B

Real and distinct

C

Real and equal

D

None of these

Answer

Real and distinct

Explanation

Solution

Given equation is 1b1 - b

Its discriminant b1α,β2x22(m2+1)x+m4+m2+1=0b - 1\alpha,\beta 2x^{2} - 2(m^{2} + 1)x + m^{4} + m^{2} + 1 = 0

which is positive, since α2+β2\alpha^{2} + \beta^{2} are real and m2m^{2}.

Hence roots are real and distinct.