Question

If $a<c<b$ then the roots of the equation $(a-b{)}^2{x}^2+2(a+b-2c)x+1=0$ are-

• A. Imaginary
• B. Real
• C. One real and one Imaginary
• D. Equal and Imaginary

Answer 1

The correct option is A Imaginary

$a<c<b$

${(a-b)}^2{x}^2+2(a+b-2c)x+1=0$

$△=4{(a+b-2c)}^2-4{(a-b)}^2$

$=4{((a-c)+(b-c))}^2-4{((a-c)-(b-c))}^2$

$=4\left(4(a-c)(b-c)\right)$

$=16(a-c)(b-c)$

$a-c<0$ & $b-c>0(\because a<c<b)$

$\therefore △<0$

$\therefore$Roots of given equation are imaginary.