Solving a quadratic equation by completing the square

Example 1: Solve $(x+9)^2-7=0$

Solution:

We have a square term containing the variable, $(x+9)^2$, and a constant term, $-7$. First, isolate the square term by adding $7$ to both sides:

$(x+9)^2=7$

Now use the square root property,

$x+9=\pm\sqrt{7}$

or

$x=-9\pm\sqrt{7}$

So, we have two values of $x$, one with the $+$ sign and the other with the $-$ sign:

That is, $x=-9+\sqrt{7}$; $x=-9-\sqrt{7}$

Example 2: Solve the equation by completing the square.

$x^2+10x-18=0$

Solution:

To complete the square, keep the variable terms on one side and move the constant term to the other side,

$x^2+10x=18$

Now square half the coefficient of $x$ and add the result to both sides of the equation. Half of $10$ is $5$, so we add $5^2$:

$x^2+10x+5^2=18+5^2$

Now, the left-hand side is in the form $a^2+2ab+b^2$, with $a=x$ and $b=5$. We can write this as $(a+b)^2$. Therefore,

$(x+5)^2=18+25$

or

$(x+5)^2=43$

Use the square root property,

$x+5=\pm \sqrt{43}$

or

$x=-5\pm \sqrt{43}$

Example 3: Solve the equation by completing the square.

$x^2-6x-8=0$

Solution:

To complete the square, keep the variable terms on one side and move the constant term to the other side,

$x^2-6x=8$

Now square half the coefficient of $x$ and add the result to both sides of the equation:

$x^2-6x+3^2=8+3^2$

Now, the left-hand side is in the form $a^2-2ab+b^2$, with $a=x$ and $b=3$. We can write this as $(a-b)^2$. Therefore,

$(x-3)^2=8+9$

or

$(x-3)^2=17$

Use the square root property,

$x-3=\pm \sqrt{17}$

or

$x=3\pm \sqrt{17}$