Draw the graph of $y = x^{2} + x - 12$ and hence solve $x^{2} + 2 x + 2 = 0$ .

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Solution

Let us first draw the graph for $y = x^{2} + x - 12$

$\begin{matrix} x & - 5 & - 4 & - 3 & 0 & 1 & 2 & 3 y & 8 & 0 & - 6 & - 12 & - 10 & - 6 & 0 \end{matrix}$

Plot the point on graph sheet

Scale: x-axis 1 cm = 5 unit; y-axis 1 cm = 5 unit

Now, by solving $y = x^{2} + x - 12$ and $x^{2} + 2 x + 2 = 0$ .

We get, y = -x -14

For the straight line y = -x -14, consider the following table

$\begin{matrix} x & - 4 & - 2 & - 1 y & - 10 & - 12 & - 15 \end{matrix}$

Scale: x-axis 1 cm = 2 unit; y-axis 1 cm = 2 unit

The straight line y = -x -14 does not intersect the parabola $y = x^{2} + x - 12$

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Draw the graph of $y = x^{2} - x - 8$ and hence, solve the equation $x^{2} - 2 x - 15 = 0$

(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.

y = x^{2}

y = x + 2

x^{2} - x - 2 = 0

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

x^{2} + 2 x + 4 = 0

y = x^{2}

y = 2 + x

x^{2} - x - 2 = 0

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