Solving quadratic equations 1

Being able to complete the square helps us to solve any quadratic equation. Here is an example. Consider the quadratic equation

x2+3x+1=0x^2+3x+1=0

Now let's complete the square for the term x2+3xx^2+3x:

x2+3x+1=0(x+32)2−(32)2+1=0(x+1.5)2−2.25+1=0(x+1.5)2−1.25=0\begin{array}{lll} x^2+3x+1&=& 0\\ \left(x+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+1 &=&0\\ \left(x+1.5\right)^2-2.25+1 &=&0\\ \left(x+1.5\right)^2-1.25 &=&0\\ \end{array}

Thus it follows

(x+1.5)2=1.25\left(x+1.5\right)^2=1.25

Taking the root on both sides, we get

x+1.5=±1.25x+1.5 = \pm \sqrt{1.25}

and thus we get the two solutions

x1,2=±1.25−1.5x_{1,2}=\pm \sqrt{1.25}-1.5

That is, we have

x1=1.25−1.5=−0.381...‾x_1=\sqrt{1.25}-1.5=\underline{-0.381...} x2=−1.25−1.5=−2.681...‾x_2=-\sqrt{1.25}-1.5=\underline{-2.681...}

What if there is a factor in front of the x2x^2-term? Well, we can first divide by this factor. Here is an example:

Example 1: Two solutions
Solution

We first divide both sides by 33, so that we have a simple term

x2−3x+2=0x^2-3x+2=0

We then complete the square as we did above:

x2−3x+2=0(x−32)2−(32)2+2=0(x−1.5)2−2.25+2=0(x−1.5)2−0.25=0∣+0.25(x−1.5)2=0.25∣xx−1.5=±0.25=±0.5\begin{array}{lll} x^2-3x+2&=&0\\ \left(x-\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2 +2 &=&0\\ \left(x-1.5\right)^2-2.25 +2 &=&0\\ \left(x-1.5\right)^2-0.25 &=&0\quad|+0.25\\ \left(x-1.5\right)^2 &=&0.25\quad|\sqrt{\phantom{x}}\\ x-1.5&=&\pm\sqrt{0.25}\\ &=&\pm 0.5\\ \end{array}

Thus, we have

x1,2=±0.5+1.5x_{1,2}=\pm0.5+1.5

that is,

x1=0.5+1.5=2‾x_1=0.5+1.5=\underline{2}x2=−0.5+1.5=1‾x_2=-0.5+1.5=\underline{1}

Solve the quadratic equation.

3x2−9x+6=03x^2-9x+6=0

by completing the square.

Notice that sometimes we have no solution. Here is an example:

Example 2: No solution
Solution

We have

x2−3x+3=0(x−32)2−(32)2+2=0(x−1.5)2−2.25+3=0(x−1.5)2+0.75=0∣−0.75(x−1.5)2=−0.75\begin{array}{lll} x^2-3x+3&=&0\\ \left(x-\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2 +2 &=&0\\ \left(x-1.5\right)^2-2.25 +3 &=&0\\ \left(x-1.5\right)^2+0.75 &=&0\quad|-0.75\\ \left(x-1.5\right)^2 &=&-0.75\\ \end{array}

And we see that the left side is always positive or 00, but the right side is negative. So there is no value for xx such that the left side equals the right side. So this equation has no solution!

Solve the quadratic equation

x2−3x+3=0x^2-3x+3=0

by completing the square.

It is also possible to have one solution. Her is an example:

Example 3: One solution
Solution

We get

x2−10x+25=0(x−5)2−52+25=0(x−5)2=0∣xx−5=0x==5‾\begin{array}{lll} x^2-10x+25&=&0\\ \left(x-5\right)^2-5^2 +25 &=&0\\ \left(x-5\right)^2 &=&0\quad|\sqrt{\phantom{x}}\\ x-5 &=&0\\ x=&=&\underline{5} \end{array}

Solve the quadratic equation

x2−10x+25=0x^2-10x+25=0

by completing the square.

Are more than two solutions possible for a quadratic equation? No. We will see why once we discuss the quadratic functions.

Exercise 1

Solve the following equation by completing the square.

  1. x2+14x+13=0x^2+14x+13=0

  2. x2+12x−12=0x^2+\frac{1}{2}x-\frac{1}{2}=0

  3. 2x2−10x+8=02x^2-10x+8=0

  4. −2x2+3x+2=0-2x^2+3x+2=0

  5. x2−18x+81=0x^2-18x+81=0

  6. x2+x+1=0x^2+x+1=0

  7. x2+bx+c=0x^2+bx+c=0

  8. ax2+bx+c=0ax^2+bx+c=0

Solution

  1. (x+7)2−49+13=0→x1=−1‾,x2=−13‾(x+7)^2-49+13=0 \rightarrow x_1=\underline{-1}, x_2=\underline{-13}

  2. (x+14)2−116−12=0→x1=12‾,x2=−1‾(x+\frac{1}{4})^2-\frac{1}{16}-\frac{1}{2}=0 \rightarrow x_1=\underline{\frac{1}{2}}, x_2=\underline{-1}

  3. Divide both sides by a=2→x2−5x+4=0→(x−2.5)2−2.52+4=0→x1=1‾,x2=4‾a=2 \rightarrow x^2-5x+4=0 \rightarrow (x-2.5)^2-2.5^2+4=0 \rightarrow x_1=\underline{1}, x_2=\underline{4}

  4. Divide both sides by a=−2→x2−1.5x−1=0→(x−0.75)2−0.752−1=0→x1=2‾,x2=−0.5‾a=-2 \rightarrow x^2-1.5x-1=0 \rightarrow (x-0.75)^2-0.75^2-1=0 \rightarrow x_1=\underline{2}, x_2=\underline{-0.5}

  5. (x−9)2−92+81=0→x1=x2=9‾(x-9)^2-9^2+81=0 \rightarrow x_1=x_2=\underline{9}

  6. (x+0.5)2−0.52+1=0→(x+0.5)2=−0.75→(x+0.5)^2-0.5^2+1=0 \rightarrow (x+0.5)^2=-0.75 \rightarrow no solution (root of a negative number)

  7. Complete the square

    (x+b2)2−(b2)2+c=0\left(x+\frac{b}{2} \right)^2 - \left(\frac{b}{2} \right)^2 +c=0 x+b2=±(b2)2−c=±b24−c=±b2−4c4=±12b2−4cx+\frac{b}{2} = \pm \sqrt{\left(\frac{b}{2} \right)^2 - c} = \pm \sqrt{\frac{b^2}{4}-c}=\pm \sqrt{\frac{b^2-4c}{4}}=\pm \frac{1}{2}\sqrt{b^2-4c}

    So we get

    x1,2=−b2±12b2−4c=−b±b2−4c2x_{1,2}=-\frac{b}{2} \pm \frac{1}{2}\sqrt{b^2-4c}=\frac{-b\pm \sqrt{b^2-4c}}{2}

  8. Divide by aa,

    x2+ba+ca=0x^2+\frac{b}{a}+\frac{c}{a}=0

    then complete the square

    (x+b2a)2−(b2a)2+ca=0\left(x+\frac{b}{2a} \right)^2 - \left(\frac{b}{2a} \right)^2 +\frac{c}{a}=0 x+b2a=±(b2a)2−ca=±b24a2−ca=±b2−4ca4a2=±12ab2−4acx+\frac{b}{2a} = \pm \sqrt{\left(\frac{b}{2a} \right)^2 - \frac{c}{a}} = \pm \sqrt{\frac{b^2}{4a^2}-\frac{c}{a}}=\pm \sqrt{\frac{b^2-4ca}{4a^2}}=\pm \frac{1}{2a}\sqrt{b^2-4ac}

    So we get

    x1,2=−b2a±12ab2−4ac=−b±b2−4ac2ax_{1,2}=-\frac{b}{2a} \pm \frac{1}{2a}\sqrt{b^2-4ac}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}