Question

Which of the following equation has two equal real toots ?

• A. $3x^2+14x-5=0$
• B. $4x^2+2x-1=0$
• C. $9x^2-6x+1=0$
• D. $x^2-5x+4=0$

Answer 1

The correct option is C $9x^2-6x+1=0$

(A) $3x^2+14x-5=0$

Then $b^2-4ac=0$ is if roots are equal

$b^2-4ac\Rightarrow (14{)}^2-4\times 3\times (-5)=196+60=256>0$

Then roots are not equal

(B) $4x^2+2x-1=0$

Then $b^2-4ac=0$ is if roots are equal

$b^2-4ac\Rightarrow (2{)}^2-4\times 4\times (-1)=4+16=20>0$

Then roots are not equal

(C) $9x^2-6x+1=0$

Then $b^2-4ac=0$ is if roots are equal

$b^2-4ac\Rightarrow (-6{)}^2-4\times 9\times (1)=36-36=0$

Then roots are equal

(D) $x^2-5x+4=0$

Then $b^2-4ac=0$ is if roots are equal

$b^2-4ac\Rightarrow (-5{)}^2-4\times 1\times (4)=25-16=9>0$

Then roots are not equal

Then option (C) $9x^2-6x+1=0$ have equal roots