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Question: If \(= 2\log(x + 1)\) are in G.P., then roots of \(x^{2} - |x + 2| + x > 0,\) are always....

If =2log⁡(x+1)= 2\log(x + 1) are in G.P., then roots of

x2−∣x+2∣+x>0,x^{2} - |x + 2| + x > 0, are always.

A

Real

B

Imaginary

C

Greater than 1

D

Equal

Answer

Real

Explanation

Solution

Given 12\frac{1}{2}

⇒ a−2b+ca - 2b + c

Now for a−2b+ca - 2b + c

Consider 1a+b−x\frac{1}{a + b - x}

Hence, roots are always real.