 Hong Kong
Stage 4 - Stage 5

# Solving by completing the square II

Lesson

We have already seen how we can use the technique of completing the square to solve many types of quadratic equations. In most of those cases our answers were whole numbers, like $x=\pm2$x=±2 or $x=3$x=3 and $x=-7$x=7, or the solutions may have been rational numbers like $x=\frac{5}{2}$x=52 and $x=\frac{-8}{11}$x=811.

An irrational number is a number that cannot be expressed as a ratio of two whole numbers. There are many quadratic equations that we will come across that will have irrational solutions. These could be in the form of surds (or radicals) like $x=\sqrt{2}$x=2 and $x=5\sqrt{13}$x=513, or in the form of non-terminating or non-repeating decimals like and $x=1.429538\dots$x=1.429538 and $x=\pi$x=π.

Luckily the method of completing the square is still the same. Let's look at an example of a quadratic equation with irrational solutions.

#### Examples

##### question 1

Solve the quadratic equation $x^2+8x=7$x2+8x=7.

Think: This quadratic looks familiar enough, but we will see that the solutions are not so simple.

Do: We can begin by completing the square on the left hand side as normal:

 $x^2+8x$x2+8x $=$= $7$7 $x^2+8x+\left(\frac{8}{2}\right)^2$x2+8x+(82​)2 $=$= $7+\left(\frac{8}{2}\right)^2$7+(82​)2 $x^2+8x+16$x2+8x+16 $=$= $7+16$7+16 $\left(x+4\right)^2$(x+4)2 $=$= $23$23 $x+4$x+4 $=$= $\pm\sqrt{23}$±√23 $x$x $=$= $-4\pm\sqrt{23}$−4±√23

So in this case our solutions are the irrational numbers $x=-4+\sqrt{23}$x=4+23 and $x=-4-\sqrt{23}$x=423. Rounded to three decimal places, the values are $x=0.796$x=0.796 and $x=-8.796$x=8.796.

##### question 2

Find the zeros of the quadratic function $y=x^2-10x+7$y=x210x+7.

Think: Remember that the zeros of a quadratic function can be found by setting the equation equal to zero, since the function will cross the $x$x-axis when $y=0$y=0.

Do: We will first let $y=0$y=0, then continue to complete the square.

 $x^2-10x+7$x2−10x+7 $=$= $0$0 $x^2-10x$x2−10x $=$= $-7$−7 $x^2-10x+\left(\frac{10}{2}\right)^2$x2−10x+(102​)2 $=$= $-7+\left(\frac{10}{2}\right)^2$−7+(102​)2 $x^2-10x+25$x2−10x+25 $=$= $-7+25$−7+25 $\left(x-5\right)^2$(x−5)2 $=$= $18$18 $x-5$x−5 $=$= $\pm\sqrt{18}$±√18

At this stage we can simplify the right hand side further by noticing that $18=2\times3^2$18=2×32.

 $x-5$x−5 $=$= $\pm\sqrt{2\times3^2}$±√2×32 $x-5$x−5 $=$= $\pm3\sqrt{2}$±3√2 $x$x $=$= $5\pm3\sqrt{2}$5±3√2

Now we have the zeros of the function, $x=5+3\sqrt{2}$x=5+32 and $x=5-3\sqrt{2}$x=532, and they are also irrational.

##### Question 3

Solve for $x$x:

$\left(x+5\right)^2-2=15$(x+5)22=15

##### Question 4

Solve for $x$x by first completing the square.

$x^2-2x-32=0$x22x32=0

##### Question 5

Solve for $x$x by first completing the square. Express your solutions in simplest fraction form.

$x^2-9x+2=0$x29x+2=0