Question

Which of the following equations has 2 as a root?

• A. $4x^2-16x+16=0$
• B. $3x^2-6x-2=0$
• C. $4x^2-16x-16=0$
• D. $4x^2-16x+5=0$

Answer 1

The correct option is A

$4x^2-16x+16=0$

(a) Substituting x=2 in $4x^2-16x+16=0$, we get $4(2{)}^2-16(2)+16$
$=16+16-32=0$
So, x=2 is a root of $4x^2-16x+16$

(b)Subsituting x=2 in $3x^2-6x-2=0$, we get $3(2{)}^2-6(2)-2$ $=12-12-2=-2\ne 0$
So, x =2 is not a root of $3x^2-6x-2=0$.

(c)Subsituting x=2 in $4x^2-16x-16=0$, we get $4(2{)}^2-16(2)-16$ $=16-32-16=-32\ne 0$
So, x =2 is not a root of $4x^2-16x-16=0$.

(d)Subsituting x=2 in $4x^2-16x+5=0$, we get $4(2{)}^2-16(2)+5$ $=16-32+5=-11\ne 0$
So, x =2 is not a root of $4x^2-16x+5=0$.