SOLVING QUADRATIC EQUATIONS BY GRAPHING EXAMPLES

Example 1 :

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

Solution :

Now, let us draw the graph of y = x² + 3x - 4

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)² + 3(-3/2) - 4

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

y = -4

Vertex (-3/2, -4)

y  =  x² + 3x - 4 ------(1)

0  =  x² + 3x - 4  ------(2)

(1) - (2)

      y  =  x² + 3x - 4 ------(1)

      0  =  x² + 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.

Example 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

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.

Example 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

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.

Example 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

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.

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