Lesson 1, Topic 1
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In order to achieve using the graphical method to solve simultaneous equations including one linear and one quadratic, we need to plot the graph of each equation on the same axes.

Then mark off the x and y coordinates of the point of intersection of both curves to obtain the required solution.

### Example 1:

(a) Copy and complete the following table of values for the relation
y = 2 + x – x2

(b) Draw the graph of the relation, using a scale of 2cm to 1 unit on each axis.

(c) Using the same axes, draw the graph of y = 1 – x.

(d) Use your curves to find the solution of the equation 1 + 2x – x2 =0 (SSCE).

Solution:

(a) y = 2 + x – x2

(b) & (c) y = 1 – x

Scale: 1unit ≡ 2cm on both axes.

(d) 1 + 2x – x2 = 0

I.e. 1 + 1 + 2x – x2 =1

⇒ 2 + 2x – x2 = 1

Subtract x from both sides

⇒      2 + 2x – x – x2 = 1 – x

⇒       2 + x – x2 = 1 – x

This suggests the intersection of the two curves y = 2 + x – x2 and y = 1 – x

From the graph x = – 0.4 and x = 2.4

### Example 2:

(a) Draw the graph of y = 6 – x – 2x2, taking the values of x from -3 to 3

(b) On the same axes, draw the graph of 3y – 5x + 6 = 0

(c) Use your graphs to find:

(i) The solution to 6 – x – 2x2 = 0
(ii) The values of x for which 3(6 – x – 2x2) = 5x – 6
(iii) The maximum value of y and the value of x for which it occurs.

(d) For what range of values of x is 6 – x – 2x2 > 0?

(e) If 6 – x – 2x2 = $$\frac {1}{3} \scriptsize (5x-6)$$ express the intersection of the graphs in the form ax2 + bx + c = 0

Solution:

(a) For y = 6 – x – 2x2

(b) 3y – 5x + 6 = 0

y = $$\frac{5x}{3} \scriptsize \:-\: 2$$

Scale: 1 Unit ≡ 1cm on x-axis

2 Units ≡ 1cm on y-axis

(c) (i) x = -2 or x = 1.5

(ii) x = – 2.8 or x = 1.4

(iii) Ymax= 6.1 at x = 0.25

(d) What range of values of x is 6 – x – 2x2 > 0?

The range of values is the part of the graph of y = 6 – x – 2x2 that y is greater than 0.

Range: -2 < x < 1.5

(e) Given 6 – x – 2x2 = $$\frac {1}{3} \scriptsize (5x-6)$$

Multiply both sides by 3

i.e. 18 – 3x – 6x2 = 5x – 6

6x2 + 8x – 24 = 0

Divide both sides by 2

⇒       3x2 + 4x – 12 = 0

⇒        3x2 + 4x – 12 ≡ ax2 + bx + c

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