Question

Question 3
Which of the following equations has 2 as a root?
(a) $x^2-4x+5=0$
(b) $x^2+3x-12=0$
(c) $2x^2-7x+6=0$
(d) $3x^2-6x-2=0$​​​​​​​

Answer 1

a) Substituting x = 2 in $x^2$ - 4x + 5, we get $(2{)}^2$ – 4(2) + 5
= 4 – 8 + 5 = 1 $\ne$ 0
So, x = 2 is not a root of $x^2$ – 4x + 5 = 0
b) Substituting x = 2 in $x^2+3x-12$, we get
$(2{)}^2$ + 3(2) – 12
= 4 + 6 – 12 = - 2 $\ne$ 0
So, x = 2 is not a root of $x^2$ + 3x – 12 = 0
c) Substituting x= 2 in $2x^2$ – 7x + 6, we get
= $2(2{)}^2$ – 7(2) +6)
= 8 – 14 +6 = 0
So, x = 2 is a root of $2x^2-7x+6=0$
d ) Substituting x = 2 in $3x^2-6x-2$ ​​​​​, we get
= $3(2{)}^2$ – 6(2) – 2
= 12 – 12 – 2 = - 2 $\ne$ 0
So, x = 2 is not a root of $3x^2-6x-2=0$​​​​​​​​​​​​​​