# SOLVING QUADRATIC EQUATIONS BY GRAPHING

Solving Quadratic Equations by Graphing :

Here we are going to see some example problems of solving quadratic equations by graphing.

## How to Solve Two Quadratic Equations by Graphing Step by Step ?

First, we have to draw the graph for the given quadratic equation. To solve the second quadratic equation using the first, we have to subtract the second equation from the first equation. So we get a straight line.

We can determine roots of a quadratic equation graphically by choosing appropriate parabola and intersecting it with a desired straight line.

(i) If the straight line intersects the parabola at two distinct points, then the x coordinates of those points will be the roots of the given quadratic equation.

(ii) If the straight line just touch the parabola at only one point, then the x coordinate of the common point will be the single root of the quadratic equation.

(iii) If the straight line doesn’t intersect or touch the parabola then the quadratic equation will have no real roots.

## Solving Quadratic Equations by Graphing - Examples

Question 1 :

Draw the graph of y = x2−4 and hence solve x2−x −12 = 0

Solution :

Now, let us draw the graph of y = x2−4

 x-4-3-2-101234 x216941014916 -4-4-4-4-4-4-4-4-4-4 y1250-3-4-30512

Points to be plotted :

(-4, 12) (-3, 5) (-2, 0) (-1, -3) (0, -4) (1, -3) (2, 0) (3, 5) (4, 12)

To find the x-coordinate of the vertex of the parabola, we may use the formula x = -b/2a

x = 0/2(2)  =  0

By applying x = 0, we get the value of y.

y = 02 - 4

y = -4

Vertex (0, -4) y  =  x2−4 ------(1)

0  =  x2−x −12  ------(2)

(1) - (2)

y  =  x− 4

0  =  x2−x −12

(-)   (-)  (+)  (+)

-----------------

y = x + 8

Now let us draw a graph for the line

 x-3-2-10123 x-3-2-10123 +8+8+8+8+8+8+8+8 y567891011

Points on line :

(-3, 5) (-2, 6) (-1, 7) (0, 8) (1, 9) (2, 10) (3, 11) The parabola and a line intersect at two points (-3, 5) and (4, 12).

Hence the solutions are -3 and 4.

Let us look into the next example on "Solving Quadratic Equations by Graphing".

Question 2 :

Draw the graph of y = x2 + x and hence solve x2 + 1 = 0

Solution :

Now, let us draw the graph of y = x2 + x

 x-4-3-2-101234 x216941014916 +x-4-3-2-101234 y126200261220

Points to be plotted :

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

To find the x-coordinate of the vertex of the parabola, we may use the formula x = -b/2a

x = 1/2(1)  =  1/2

By applying x = 1/2, we get the value of y.

y = (1/2)2 + (1/2)

=  (1/4) + (1/2)

=  (1 + 2)/4

=  3/4

Vertex (1/2, 3/4) y  =  x+ x ------(1)

0  =  x2 + 1  ------(2)

(1) - (2)

y  =  x2 + x

0  =  x2 + 1

(-)   (-)  (-)

-----------------

y = x - 1

Now let us draw a graph for the line Points on line :

(-3, -4) (-2, -3) (-1, -2) (0, -1) (1, 0) (2, 1) (3, 2) The parabola and a line do not intersect at anyware. So it has no real roots.

Let us look into the next example on "Solving Quadratic Equations by Graphing".

Question 3 :

Draw the graph of y = x2 + 3x + 2 and use it to solve x2 + 2x + 1 = 0.

Solution :

Now let us draw the graph of y = x2 + 3x + 2

 x-4-3-2-101234 x216941014916 +3x-12-9-6-3036912 +2+2+2+2+2+2+2+2+2+2 y620026122030

Points to be plotted :

(-4, 6) (-3, 2) (-2, 0) (-1, 0) (0, 2) (1, 6) (2, 12) (3, 20) (4, 30)

To find the x-coordinate of the vertex of the parabola, we may use the formula x = -b/2a

x = -3/2(1)  =  -3/2

By applying x = -3/2, we get the value of y.

y = x2 + 3x + 2

=  (-3/2)2 + 3(-3/2) + 2

=  (9/4) - (9/4) + 2

=  2

Vertex (-3/2, 2) y  =  x2 + 3x + 2 ------(1)

0  =  x2 + 2x + 1 ------(2)

(1) - (2)

y  =  x2 + 3x + 2

0  =  x2 + 2x + 1

(-)   (-)   (-)    (-)

--------------------

y = x + 1

Now let us draw a graph for the line points on the line :

(-3, -2) (-2, -1) (-1, 0) (0, 1) (1, 2) (2, 3) (3, 4) The line and the parabola intersects at one point, that is -1.

Hence -1 is the solution. After having gone through the stuff given above, we hope that the students would have understood, "Solving Quadratic Equations by Graphing".

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