**Linear Quadratic Systems :**

A linear quadratic system is a system containing one linear equation and one quadratic equation which may be one straight line and one parabola*, *or one straight line and one circle.

**Example 1 : **

Solve y = 2x² and y = - x + 6 graphically.

**Solution : **

First let us make a table of values to graph y = 2x²

We can get the following points from the table.

(–3, 18), (–2, 8),(–1,2), (0, 0), (1, 2), (2, 8), (3, 18) ----- (1)

Now, let us make a table of values to graph y = -x + 6

We can get the following points from the table.

(–1, 7), (0, 6), (1, 5), (1, 5), (2, 4) ----- (2)

Plotting the points which we have in (1) and (2), we get the graph of y = 2x² and y = -x + 6

From the graph, the points of intersection or the two solutions for the given system are

(-2, 8) and (1.5, 4.5)

**Example 2 : **

Solve y = x² + 3x +2 and y = x - 2 graphically.

**Solution : **

First let us make a table of values to graph y = x² + 3x +2

We can get the following points from the table.

(–4, 6), (–3, 2), (–2, 0), (–1, 0), (0, 2), (1, 6), (2, 12) and (3, 20) ----- (1)

Now, let us make a table of values to graph y = x - 2

We can get the following points from the table.

(-2, -4), (0, -2), (1, -1), (2, 0) ----- (2)

Plotting the points which we have in (1) and (2), we get the graph of y = x² + 3x +2 and y = x - 2

In the above graph, the straight line y = x - 2 does not intersect y = x² + 3x +2.

Hence, there is no solution for the given system.

**Example 3 : **

Draw the graph of y = 2x² and hence solve 2x²+x-6 = 0.

**Solution : **

First let us make a table of values to graph y = 2x²

We can get the following points from the table.

(–3, 18), (–2, 8),(–1,2), (0, 0), (1, 2), (2, 8), (3, 18) ----- (1)

Now, let us take the quadratic equation 2x²+x-6 = 0.

Form the first equation, we know y = 2x².

So, plugging 2x² = y in (2x²+x-6 = 0), we get

y + x - 6 = 0

y = -x + 6

Now, let us make a table of values to graph y = -x + 6

We can get the following points from the table.

(–1, 7), (0, 6), (1, 5), (1, 5), (2, 4) ----- (2)

Thus, the roots of 2x² + x - 6 = 0 are nothing but the x - coordinates of point of intersection of y = 2x² and y = -x + 6.

Plotting the points which we have in (1) and (2), we get the graph of y = 2x² and y = -x + 6

In the graph above, the points of intersection of the line and the parabola are

(-2, 8) and (1.5, 4.5)

The x-coordinates in the points of intersection are -2 and 1.5.

Hence, the two solutions of the equation 2x²+x-6 = 0 are

-2 and 1.5

**Example 4 : **

Draw the graph of y = x² + 3x + 2 and use it to solve the equation x² + 2x + 4 = 0.

**Solution : **

First let us make a table of values to graph y = x² + 3x + 2

We can get the following points from the table.

(–4, 6), (–3, 2), (–2, 0), (–1, 0), (0, 2), (1, 6), (2, 12) and (3, 20) ----- (1)

Now, let us take the quadratic equation x²+ 2x + 4 = 0

x²+ 2x + 4 = 0

x²+ (3x-x) + (2+2) = 0

x²+ 3x + 2 - x + 2 = 0

Form the first equation, we know y = x² + 3x + 2.

So, plugging x² + 3x + 2 = y in (x²+ 3x + 2 - x + 2 = 0), we get

y - x + 2 = 0

y = x - 2

Now, let us make a table of values to graph y = x - 2

We can get the following points from the table.

(-2, -4), (0, -2), (1, -1), (2, 0) ----- (2)

Thus, the roots of x² + 2x + 4 = 0 are obtained from the points of intersection of y = x - 2 and y = x² + 3x + 2

Plotting the points which we have in (1) and (2), we get the graph of y = x² + 3x +2 and y = x - 2

In the above graph, the straight line y = x - 2 does not intersect y = x² + 3x +2.

Hence, the quadratic equation x² + x + 4 = 0 has no real roots.

After having gone through the stuff given above, we hope that the students would have understood "Linear quadratic systems".

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