Question

In each of the examples given below determine whether the values given against each of the quadratic equation are the roots of the equation or not.
y² + 5y + 6 = 0, y = 4, – 2, –3

Answer 1

Given: y² + 5y + 6 = 0 and y = 4, – 2, –3
We have to determine whether the values of y satisfy the quadratic equation .

On substituting y = 4 in L.H.S. of the given equation, we get:
(4)² + 5(4) + 6
= 16 + 20 + 6
= 42
=> L.H.S. $\ne$ R.H.S.
Thus, y = 4 does not satisfy the given equation.
Therefore, y = 4 is not a root of the given quadratic equation.

On substituting y = –2 in L.H.S. of the given equation, we get:
(–2)² + 5(–2) + 6
= 4 – 10 + 6
= 0
=>L.H.S. = R.H.S.
Thus, y = –2 satisfies the given equation.
Therefore, y = –2 is a root of the given quadratic equation.

On substituting y = –3 in L.H.S. of the given equation, we get:
(–3)² + 5(–3) + 6
= 9 – 15 + 6
=0
=> L.H.S. = R.H.S.
Thus, y = –3 satisfies the given equation.
Therefore, y = –3 is a root of the given quadratic equation.