SOLVING QUADRATIC EQUATIONS BY GRAPHING EXAMPLES

Solving Quadratic Equations by Graphing Examples :

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.

Let us see the some examples on "Solving Quadratic Equations by Graphing Examples".

Solving Quadratic Equations by Graphing Examples - Questions

Question 1 :

Draw the graph of y = x+ 3x - 4 and hence solve x2 + 3x - 4 = 0

Solution :

Now, let us draw the graph of y = x+ 3x - 4

x

-4

-3

-2

-1

0

1

2

3

4

x2

16

9

4

1

0

1

4

9

16

3x

-12

-9

-8

-3

0

3

6

9

12

-4

-4

-4

-4

-4

-4

-4

-4

-4

-4

y

0

-4

-8

-6

-4

0

6

14

24

Points to be plotted :

(-4, 0) (-3, -4) (-2, -8) (-1, -6) (0, -4) (1, 0) (2, 6) (3, 14) (4, 24)

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 = (-3/2)+ 3(-3/2) - 4

y = 9/4 - (9/4) - 4

y = -4

Vertex (-3/2, -4)

y  =  x+ 3x - 4 ------(1)

0  =  x+ 3x - 4  ------(2)

(1) - (2)

      y  =  x+ 3x - 4 ------(1)

      0  =  x+ 3x - 4  ------(2)

    (-)     (-)   (-)  (-)

    -------------------

       y = 0 

y = 0 means x-axis. The parabola intersects the x-axis at two points-4 and 1. 

Hence the solutions are -4 and 1.

Let us see the next example on "Solving Quadratic Equations by Graphing Examples".

Question 2 :

Draw the graph of y = x2 −5x −6 and hence solve x2 −5x −14 = 0

Solution :

Now, let us draw the graph of y = x2 - 5x - 6

x

-4

-3

-2

-1

0

1

2

3

4

x2

16

9

4

1

0

1

4

9

16

-5x

20

15

10

5

0

-5

-10

-15

-20

-6

-6

-6

-6

-6

-6

-6

-6

-6

-6

y

30

18

8

0

-6

-10

-12

-12

-10

Points to be plotted :

(-4, 30) (-3, 18) (-2, 8) (-1, 0) (0, -6) (1, -10) (2, -12) (3, -12) (4, -10)

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

x = -(-5)/2(1)  =  5/2

By applying x = 5/2, we get the value of y.

y = (5/2)2 - 5(5/2) - 6

y = 25/4 - (25/2) - 6

y = (25 - 50 - 24)/4

y = -49/4

Vertex (5/2, -49/4)

y  =  x2 −5x −6 ------(1)

0  =  x2 −5x −14  ------(2)

(1) - (2)

     y  =  x2 −5x −6 ------(1)

     0  =  x2 −5x −14  ------(2)

    (-)     (-)  (+)  (+)

    -------------------

       y = 8 

Hence the parabola and the line intersect at two points -2 and 7 on the x-axis.

Hence the solutions are -2 and 7.

Let us see the next example on "Solving Quadratic Equations by Graphing Examples".

Question 3 :

Draw the graph of y = 2x2 − 3x − 5 and hence solve 2x2 − 4x − 6 = 0

Solution :

Let us draw the graph of y = 2x2 − 3x − 5

x

-2

-1

0

1

2

3

4

2x2

8

2

0

2

8

18

32

-3x

6

3

0

-3

-6

-9

-12

-5

-5

-5

-5

-5

-5

-5

-5

y

9

0

-5

-6

-3

4

15

Points to be plotted :

 (-2, 9) (-1, 0) (0, -5) (1, -6) (2, -3) (3, 4) (4, 15)

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

x = -(-3)/2(2)  =  3/4

By applying x = 3/4, we get the value of y.

y = 2(3/4)2 − 3(3/4) − 5

y = 18/16 - (12/4) - 5

y = (18 - 48 - 80)/16

y = -110/16

Vertex (3/4, -110/16)

y  =  2x2 − 3x − 5 ------(1)

0  =  2x2 − 4x − 6  ------(2)

(1) - (2)

      y  =  2x2 − 3x − 5 ------(1)

      0  =  2x2 − 4x − 6  ------(2)

      (-)     (-)  (+)  (+)

    -------------------

       y  =  x + 1 

By drawing a line in the same graph, we get

The parabola and a line intersect at two points -1 and 3. Hence the solutions are -1 and 3.

Let us see the next example on "Solving Quadratic Equations by Graphing Examples".

Question 4 :

Draw the graph of y = (x − 1)(x + 3) and hence solve x2 − x − 6 = 0

Solution :

y = (x − 1)(x + 3) 

y = x2 + 3x - x - 3

y = x2 + 2x - 3

x

-2

-1

0

1

2

3

4

x2

4

1

0

1

4

9

16

2x

-4

-2

0

2

4

6

8

-3

-3

-3

-3

-3

-3

-3

-3

y

-3

-4

-3

0

5

12

21

Points to be plotted :

 (-2, -3) (-1, -4) (0, -3) (1, 0) (2, 5) (3, 12) (4, 21)

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

x = -2/2(1)  =  -1

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

y = (-1)2 + 2(-1) - 3

y = 1 - 2 - 3

y = -4

Vertex (-1, -4)

y  =   x2 + 2x - 3 ------(1)

0  =  x2 − x − 6  ------(2)

(1) - (2)

     y  =   x2 + 2x - 3 ------(1)

      0  =  x2 − x − 6  ------(2)

      (-)    (-)  (+)  (+)

    -------------------

       y  =  3x + 3

 y = 3(x + 1)

By drawing a line in the same graph, we get

The parabola and a line intersect at two points -2 and 7. Hence the solutions are -2 and 7.

After having gone through the stuff given above, we hope that the students would have understood, "Solving Quadratic Equations by Graphing Examples". 

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