**Linear Quadratic Systems Worksheet :**

Worksheet on linear quadratic systems is much useful to the students who would like to practice problems on the system of equations where one equation is linear and the other one is quadratic.

**Problem 1 : **

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

**Problem 2 :**

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

**Problem 3 :**

Draw the graph of y = 2x² and hence solve

2x² + x - 6 = 0

**Problem 4 :**

Draw the graph of y = x²+ 3x + 2 and hence solve

x² + 2x + 4 = 0

**Problem 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)

**Problem 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.

**Problem 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

**Problem 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 worksheet".

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