Question

x² – 7x – 18 = 0, x =$\sqrt{3}$ , – 2, 4

Answer 1

We are given a quadratic equation x² – 7x – 18 = 0 and the values of x as x =$\sqrt{3}$ , – 2, 4
we have to determine if the given values of x satisfy the quadratic equation

On substituting x =$\sqrt{3}$ in L.H.S. of the given equation, we get
L.H.S. = $(\sqrt{3}{)}^2-7(\sqrt{3})-18$
= 3 – 7$\sqrt{3}$ – 18
= –15 – 7$\sqrt{3}$
=> L.H.S. $\ne$ R.H.S.
Thus, x = $\sqrt{3}$ does not satisfy the given equation.
Therefore, x =$\sqrt{3}$ is not a root of the given quadratic equation.

On substituting x = –2 in L.H.S. of the given equation, we get
L.H.S. = (–2)² – 7(–2) – 18
= 4 + 14 – 18
= 0
=>L.H.S. = R.H.S.
Thus, x = –2 satisfies the given equation.
Therefore, x = –2 is a root of the given quadratic equation.

On substituting x = 4 in L.H.S. of the given equation, we get
L.H.S. = (4)² – 7(4) – 18
= 16 – 28 – 18
= –30
=> L.H.S. $\ne$ R.H.S.
Thus, x = 4 does not satisfy the given equation.
Therefore, x = 4 is not a root of the given quadratic equation.