Draw the graph of $y = 2 x^{2}$ and hence solve $2 x^{2} + x - 6 = 0$

Open in App

Solution

$y = 2 x^{2}$
$x$ $- 3$ $- 2$ $- 1$ $0$ $1$ $2$ $3$
$x^{2}$ $9$ $4$ $1$ $0$ $1$ $4$ $9$
$y = 2 x^{2}$ $18$ $8$ $2$ $0$ $2$ $8$ $18$

The points are $(- 3, 18), (- 2, 8), (- 1, 2), (0, 0), (1, 2), (2, 8), (3, 18)$

$y = 2 x^{2}, 2 x^{2} + x - 6 = 0$
$y + x - 6 = 0$
$y = - x + 6$
$x$ $- 1$ $0$ $1$ $2$
$- x$ $1$ $0$ $- 1$ $- 2$
$6$ $6$ $6$ $6$ $6$
$7$ $6$ $5$ $4$
The points are $(- 1, 7), (0, 6), (1, 5), (2, 4)$
The solution set is ${- 2, 1.5}$

Suggest Corrections

Similar questions

y = 2 x^{2}

2 x^{2} + x - 6 = 0

(1) Draw the graph of y = x² and y = 2x + 3 and hence solve the equation x² − 2x − 3 = 0.

(2) Draw the graph of y = 2x²and y = 3 − x and hence solve the equation 2x² + x − 3 = 0.

(3) Draw the graph of y = 2x² and y = 3 + x and hence solve the equation 2x² − x − 3 = 0.

Draw the graph $y = x^{2} - 9$ and hence solve the equation $x^{2} - 2 x - 8 = 0$

y = 2 x^{2} - x - 3

\frac{1}{x + y} - \frac{1}{2 x} = \frac{1}{30}, \frac{5}{x + y} + \frac{1}{x} = \frac{4}{3}

2 x^{2} - y^{2}

Join BYJU'S Learning Program