Second AlmightySolve Any Quadratic Equation: 3 Methods With Worked Examples
Every quadratic equation can be solved in three ways, and one of them never fails. Here are all three, one fully worked example each, then a quick guide to picking the right one.
What is a quadratic equation?
A quadratic equation has the form
\[ ax^2+bx+c=0,\qquad a\ne 0. \]It has at most two solutions, called roots. Solving it means finding every \( x \) that makes the left side equal to zero.
Method 1: Factorising
Fast, but it only works neatly when the numbers are friendly. Look for two numbers whose product is \( a\times c \) and whose sum is \( b \).
\[ x^2-5x+6=0 \]
Here \( a\times c=6 \) and \( b=-5 \). The numbers that multiply to 6 and add to \( -5 \) are \( -2 \) and \( -3 \). Split the middle term:
\[ x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)=0. \]A product is zero only if a factor is zero, so \( x-2=0 \) or \( x-3=0 \).
The roots are x = 2 and x = 3.
Check one: \( 2^2-5\cdot2+6=4-10+6=0 \). ✓
Method 2: The quadratic formula
It works for every quadratic, even when factorising looks impossible.
\[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \]The discriminant decides everything
The part under the root, \( D=b^2-4ac \), tells you how many real roots there are. Compute it first.
| D | Roots | Example |
|---|---|---|
| positive | two different real roots | \( x^2-4x+3=0 \): \( D=4 \) |
| zero | one repeated root | \( x^2-4x+4=0 \): \( D=0 \) |
| negative | no real roots | \( x^2+2x+5=0 \): \( D=-16 \) |
\[ 2x^2+5x-3=0 \]
Here \( a=2,\ b=5,\ c=-3 \). The discriminant is \( 25-4\cdot2\cdot(-3)=25+24=49 \), and \( \sqrt{49}=7 \). So
\[ x=\frac{-5\pm7}{4}:\qquad x=\frac{2}{4}=\frac12\quad\text{or}\quad x=\frac{-12}{4}=-3. \]The roots are x = 1/2 and x = −3.
A quick check you can always use
The roots of \( ax^2+bx+c=0 \) add up to \( -\tfrac ba \) and multiply to \( \tfrac ca \). Here \( -\tfrac ba=-\tfrac52 \) and \( \tfrac ca=-\tfrac32 \). The roots give \( \tfrac12+(-3)=-\tfrac52 \) and \( \tfrac12\cdot(-3)=-\tfrac32 \). Both match. ✓
Method 3: Completing the square
Slower, but it shows where the formula comes from. Turn one side into a perfect square, then take square roots.
\[ x^2+6x-7=0 \]
Move the constant across: \( x^2+6x=7 \). Take half of the coefficient of \( x \), which is 3, and square it to get 9. Add 9 to both sides:
\[ x^2+6x+9=16\quad\Longrightarrow\quad (x+3)^2=16. \]Take square roots: \( x+3=\pm4 \).
The roots are x = 1 and x = −7.
Check: \( 1+6-7=0 \). ✓
Where the formula comes from
Do the same steps on \( ax^2+bx+c=0 \). Divide by \( a \): \( x^2+\tfrac bax=-\tfrac ca \). Add \( \left(\tfrac{b}{2a}\right)^2 \) to both sides:
\[ \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\quad\Longrightarrow\quad x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \]So the quadratic formula is just completing the square, done once and for all.
Which method should you pick?
- Try factorising first, because it is quick.
- If the numbers are not friendly, use the formula. It never fails.
- Use completing the square when the question asks for it, or when you need the vertex of a parabola.
Two mistakes that cost marks
- Forgetting the minus in front of b. In Example 2, using \( +5 \) instead of \( -5 \) gives \( \tfrac{5\pm7}{4} \), which is 3 and \( -\tfrac12 \). Test 3: \( 2\cdot9+15-3=30\ne0 \), so it is wrong.
- Forgetting the ±. A positive discriminant means two roots. Write both.
Always put your roots back into the original equation to check them.
Your turn
\[ x^2-7x+12=0 \]
Solve it with any method. Which one did you choose?
Show the answer
Two numbers that multiply to 12 and add to \( -7 \) are \( -3 \) and \( -4 \), so \( (x-3)(x-4)=0 \).
The roots are x = 3 and x = 4.
Check with the sum and product: \( -\tfrac ba=7=3+4 \) and \( \tfrac ca=12=3\cdot4 \). ✓