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Quadratic Equations

Quadratic equations

Consider the following problem.

A bathroom wall has a rectangular shape and will be covered with square tiles. The length of the wall is 5 tiles longer than its width. If 300 tiles are needed to cover the wall, determine the length of the wall. Assume the area of one tile is 1 square unit.

Problems like this can be modeled using a quadratic equation.

A quadratic equation is an equation containing a variable with highest degree two.

The general form is:

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

where:

  • aa, bb, and cc are real numbers,
  • aa is the coefficient of x2x^2,
  • bb is the coefficient of xx,
  • cc is a constant,
  • a0a \neq 0.

Observe the following examples.

  1. 2x26x+5=02x^2-6x+5=0
  2. x24=0x^2-4=0
  3. x29x=0x^2-9x=0

Explanation:

  1. The equation 2x26x+5=02x^2-6x+5=0 has a=2a=2, b=6b=-6, and c=5c=5. Since all terms are present and nonzero, it is called a complete quadratic equation.
  2. The equation x24=0x^2-4=0 has a=1a=1, b=0b=0, and c=4c=-4. Since it has no xx term, it is called a pure quadratic equation.
  3. The equation x29x=0x^2-9x=0 has a=1a=1, b=9b=-9, and c=0c=0. Since its constant term is zero, it is called an incomplete quadratic equation.

Some equations are not written directly in general form, for example:

  1. x2+7x=10x^2+7x=-10
  2. 1x+1x1=32\dfrac{1}{x}+\dfrac{1}{x-1}=\dfrac{3}{2}

Such equations can still be transformed into the general quadratic form using algebraic operations.

Rewrite 2x2=3x+202x^2=3x+20 into general form, then determine aa, bb, and cc.

2x2=3x+202x23x=202x23x20=0 \begin{aligned} 2x^2 &= 3x+20 \\ 2x^2-3x &= 20 \\ 2x^2-3x-20 &= 0 \end{aligned}

Therefore:

  • a=2a=2
  • b=3b=-3
  • c=20c=-20

A value of xx that makes the equation true is called a root or solution of the quadratic equation.

In general, the roots can be found by:

  1. factoring,
  2. completing the square,
  3. using the quadratic formula.

Factoring changes an algebraic expression into a product form.

Find the roots of:

4x225=0 4x^2-25=0

Solution:

4x225=0(2x)252=0(2x5)(2x+5)=0 \begin{aligned} 4x^2-25 &= 0 \\ (2x)^2-5^2 &= 0 \\ (2x-5)(2x+5) &= 0 \end{aligned}

So:

2x5=0or2x+5=02x=5or2x=5x=52orx=52 \begin{aligned} 2x-5 &= 0 \quad \text{or} \quad 2x+5=0 \\ 2x &= 5 \quad \text{or} \quad 2x=-5 \\ x &= \frac{5}{2} \quad \text{or} \quad x=-\frac{5}{2} \end{aligned}

Thus, the roots are:

x=52orx=52 x=\frac{5}{2} \quad \text{or} \quad x=-\frac{5}{2}

Find the solution set of:

3x27x=0 3x^2-7x=0

Solution:

3x27x=0x(3x7)=0 \begin{aligned} 3x^2-7x &= 0 \\ x(3x-7) &= 0 \end{aligned}

Hence:

x=0or3x7=0x=0or3x=7x=0orx=73 \begin{aligned} x &= 0 \quad \text{or} \quad 3x-7=0 \\ x &= 0 \quad \text{or} \quad 3x=7 \\ x &= 0 \quad \text{or} \quad x=\frac{7}{3} \end{aligned}

So, the solution set is:

{0,73} \left\{0,\frac{7}{3}\right\}

Find the roots of:

x27x+10=0 x^2-7x+10=0

Find two numbers whose sum is 7-7 and whose product is 1010, namely 5-5 and 2-2.

x27x+10=0(x5)(x2)=0 \begin{aligned} x^2-7x+10 &= 0 \\ (x-5)(x-2) &= 0 \end{aligned}

Thus:

x5=0orx2=0x=5orx=2 \begin{aligned} x-5 &= 0 \quad \text{or} \quad x-2=0 \\ x &= 5 \quad \text{or} \quad x=2 \end{aligned}

So, the roots are:

x=5orx=2 x=5 \quad \text{or} \quad x=2
  1. Quadratic Function
  2. Solving Quadratic Equations
  3. Assessment of Quadratic Equations

Not all quadratic equations are easy to solve by factoring. Another method is completing the square.

Steps:

  1. make sure the coefficient of x2x^2 is 1,
  2. move the constant to the right side if needed,
  3. add the square of half the coefficient of xx to both sides,
  4. rewrite into perfect-square form,
  5. solve using square roots.

Solve:

x26x+8=0 x^2-6x+8=0

Solution:

x26x+8=0    x26x=8 x^2-6x+8=0 \iff x^2-6x=-8

Half of the coefficient of xx is:

62=3 \frac{-6}{2}=-3

Its square is:

(3)2=9 (-3)^2=9

Add 9 to both sides:

x26x=8x26x+9=8+9(x3)2=1 \begin{aligned} x^2-6x &= -8 \\ x^2-6x+9 &= -8+9 \\ (x-3)^2 &= 1 \end{aligned}

Then:

x3=±1x=3±1 \begin{aligned} x-3 &= \pm 1 \\ x &= 3 \pm 1 \end{aligned}

So:

x=4orx=2 x=4 \quad \text{or} \quad x=2

Therefore, the solution set is:

{2,4} \{2,4\}

Besides factoring and completing the square, a quadratic equation can be solved using the quadratic formula:

x1,2=b±b24ac2a x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

It is derived as follows:

ax2+bx+c=0ax2+bx=cx2+bax=ca \begin{aligned} ax^2+bx+c &= 0 \\ ax^2+bx &= -c \\ x^2+\frac{b}{a}x &= -\frac{c}{a} \end{aligned}

Add the following to both sides:

(12ba)2=(b2a)2 \left(\frac{1}{2}\cdot \frac{b}{a}\right)^2=\left(\frac{b}{2a}\right)^2

Then:

x2+bax+(b2a)2=ca+(b2a)2(x+b2a)2=ca+b24a2(x+b2a)2=4ac+b24a2(x+b2a)2=b24ac4a2x+b2a=±b24ac4a2x=b2a±b24ac4a2x=b±b24ac2a \begin{aligned} x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2 &= -\frac{c}{a}+\left(\frac{b}{2a}\right)^2 \\ \left(x+\frac{b}{2a}\right)^2 &= -\frac{c}{a}+\frac{b^2}{4a^2} \\ \left(x+\frac{b}{2a}\right)^2 &= \frac{-4ac+b^2}{4a^2} \\ \left(x+\frac{b}{2a}\right)^2 &= \frac{b^2-4ac}{4a^2} \\ x+\frac{b}{2a} &= \pm \sqrt{\frac{b^2-4ac}{4a^2}} \\ x &= -\frac{b}{2a}\pm \sqrt{\frac{b^2-4ac}{4a^2}} \\ x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{aligned}

The quantity b24acb^2-4ac is called the discriminant, denoted by DD.

Find the roots of:

x2x6=0 x^2-x-6=0

Known:

  • a=1a=1
  • b=1b=-1
  • c=6c=-6

Use the quadratic formula:

x=b±b24ac2ax=(1)±(1)24(1)(6)2(1)x=1±1+242x=1±252x=1±52 \begin{aligned} x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x &= \frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-6)}}{2(1)} \\ x &= \frac{1\pm\sqrt{1+24}}{2} \\ x &= \frac{1\pm\sqrt{25}}{2} \\ x &= \frac{1\pm5}{2} \end{aligned}

Therefore:

x1=1+52=3x2=152=2 \begin{aligned} x_1 &= \frac{1+5}{2}=3 \\ x_2 &= \frac{1-5}{2}=-2 \end{aligned}

So, the roots are:

x=3andx=2 x=3 \quad \text{and} \quad x=-2

After studying this material, complete the evaluation by following these steps:

  1. click Start Quiz,
  2. choose the correct answer,
  3. click Next for the next question,
  4. click Prev to return to the previous question,
  5. click Reset to start again,
  6. click Submit after finishing all questions,
  7. check the score displayed at the bottom,
  8. final value equals total score divided by maximum score, then multiplied by 100.

That concludes this material on quadratic equations for Grade IX SMP/MTs semester 1. If you find any mistakes, please leave a correction in the comments.

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