In each of the following, determine whether the given values are solutions (roots) of the equation or not :
(i) $3 x^{2} - 2 x - 1$ = 0; x = 1
(ii) $x^{2} + 6 x + 5$ = 0; x = – 1, x = – 5
(iii) $x^{2} + x - 4 = 0$ ; x = – 2

Open in App

Solution

(i) Step 1:- Value of $3 x^{2} - 2 x - 1$ at x = 1 is $3 (1)^{2} - 2 (1) - 1$ = 3 – 2 – 1 = 0

Step 2:- x = 1 is a solution of the given equation.

(ii) Step 1:- For x = – 1, L.H.S. = $(- 1)^{2} + 6 (- 1) + 5$ = 1 – 6 + 5 = 0 = R.H.S.

Step 2: x = – 1 is a solution of the given equation

Step 1:- For x = – 5, L.H.S. = $(- 5)^{2} + 6 (- 5) + 5$ = 25 – 30 + 5 = 0 = R.H.S.

Step 2: x = – 5 is a solution of the given equation.

iii) Step 1: For x = -2,

L.H.S. $x^{2} + x - 4 = (- 2)^{2} + (- 2) - 4$ = 4 - 2 – 4 = -2 ≠ R.H.S.

Step 2:- x =-2 is not a solution of the given equation

Suggest Corrections

Similar questions

x^{2}

x^{2} - 3 x + 2 = 0, x = 2, x = - 1

x^{2} + x + 1 = 0, x = 0, x = 1

x^{2} - 3 \sqrt{3} x + 6 = 0, x = \sqrt{3}, x = - 2 \sqrt{3}

x + \frac{1}{x} = \frac{13}{6}, x = \frac{5}{6}, x = \frac{4}{3}

2 x^{2} - x + 9 = x^{2} + 4 x + 3, x = 2, x = 3

x^{2} - \sqrt{2} x - 4 = 0, x = - \sqrt{2}, x = - 2 \sqrt{2}

a^{2} x^{2} - 3 a b x + 2 b^{2} = 0, x = \frac{a}{b}, x = \frac{b}{a}

x^{2} - 3 x + 2 = 0, x = 2, x = - 1

x^{2} - \sqrt{2} x - 4 = 0; x = - \sqrt{2}, x = - 2 \sqrt{2}

x^{2} + \sqrt{2} x - 4 = 0, x = \sqrt{2}, x = - 2 \sqrt{2}

Join BYJU'S Learning Program