Question:
Which of the following equations has 2 as a root?
(a) x2-4x + 5=0
(b) x2 + 3x-12 =0
(c) 2x2 – 7x + 6 = 0
(d) 3x2 – 6x – 2 = 0
Solution:
(c)
(a) Substitutina $x=2$ in $x^{2}-4 x+5$, we get $(2)^{2}-4(2)+5$
$=4-8+5=1 \neq 0$
So, $x=2$ is not a root of $x^{2}-4 x+5=0$
(b) Substituting $x=2$ in $x^{2}+3 x-12$, we get
$(2)^{2}+3(2)-12$
$=4+6-12=-2 \neq 0$
So, $x=2$ is not a root of $x^{2}+3 x-12=0$.
(c) Substituting $x=2$ in $2 x^{2}-7 x+6$, we get
$2(2)^{2}-7(2)+6=2(4)-14+6$
$=8-14+6=14-14=0$
So, $x=2$ is root of the equation $2 x^{2}-7 x+6=0$
(d) Substituting $x=2$ in $3 x^{2}-6 x-2$, we get
$3(2)^{2}-6(2)-2$
$=12-12-2=-2 \neq 0$
So, $x=2$ is not a root of $3 x^{2}-6 x-2=0$