Substituting values of x

For many quadratic equations they can actually be solved by substituting values of x into the equation and then drawing the graph.

If you have time our advice is:

Always do a quick sketch

Example 1

Solve x^2+3x-4=y

 x=2 2^2+3times2-4 = 4+6-4 =6 x=1 1^2+3times1-4 = 1+3-4 =0 x=0 0^2+3times0-4 = 0+0-4 =-4 x=-1 (-1)^2+3times(-1)-4 = 1-3-4 =-6 x=-2 (-2)^2+3times(-2)-4 = 4-6-4 =-6 x=-3 (-3)^2+3times(-3)-4 = 9-9-4 =-4 x=-4 (-4)^2+3times(-4)-4 = 16-12-4 =0 x=-5 (-5)^2+3times(-5)-4 = 25-15-4 =6

If we plot these coordinates on a graph we get:

Now joining up the points we get:

Here we can see that the roots are +1 and -4

So why do we not use this way always?

Because you might get a beast like:

2x^2-5x-6=y

Try plotting this one:

 x=4 2times4^2-5times4-6 = 32-20-6 =6 x=3 2times3^2-5times3-6 = 18-15-6 =-3 x=2 2times2^2-5times2-6 = 8-10-6 =-8 x=1 2times1^2-5times1-6 = 2-5-6 =-9 x=0 2times0^2-5times0-6 = 0-0-6 =-6 x=-1 2times(-1)^2-5times(-1)-6 = 2+10-6 =6

If we plot these coordinates on a graph we get:

Now joining up the points we get:

You can see that you can not find the roots graphically.

But you can find them using the quadratic formula.

Example 2

Solve 2x^2-5x-6=y  using the quadratic formula.

x=(-b+-sqrt(b\ ^2-4ac))/(2a)

In this case a=2, b=-5  and c=-6

x=(-(-5)+-sqrt((-5)^2-4times2times(-6)))/(2times2)

x=(5+-sqrt(25+48))/4

x=(5+-sqrt73)/4

x=(5+sqrt73)/4 \ \ \ or \ \ \ x=(5-sqrt73)/4

x=3.386 \ \ \ or \ \ \ x=-0.886

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