2.7 - Quadratic formula and completing the square

Learn two reliable ways to solve quadratics when factorisation is inconvenient. You will keep both roots, recognise when real roots exist and give exact or rounded answers as requested.

Using the quadratic formula (Higher)

Can $2x^2+3x-4=0$ have exact answers even when you cannot find tidy factors? Yes: a square root can be part of an exact number. We will find both answers without guessing decimal values.

You already know

A quadratic can be written as $ax^2+bx+c=0$ , with $a\ne0$ . Its coefficients include their signs. A square root such as $\sqrt{41}$ is exact; a rounded decimal is an approximation.

Bring every term to the same side before reading the coefficients. For $2x^2+3x-4=0$ , they are $a=2$ , $b=3$ , $c=-4$ . If a term is absent, its coefficient is zero.

Quadratic formula

The solutions of $ax^2+bx+c=0$ , where $a\ne0$ , are

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

The $\pm$ means do one calculation with addition and one with subtraction. The denominator divides the whole numerator. Real answers require $b^2-4ac\ge0$ .
In the Higher formulae sheet.

Why does this formula work? First rearrange to $ax^2+bx=-c$ . Multiply both sides by $4a$ , then add $b^2$ to both sides. The left side becomes $(2ax+b)^2$ , giving $(2ax+b)^2=b^2-4ac$ . Taking both square roots and solving the resulting linear equations gives the formula. We will use this same square idea more directly in the next part.

Worked example 1: exact surd roots

Question. Solve $2x^2+3x-4=0$ exactly.

Plan. Substitute the signed coefficients into the formula and keep the root exact.

\begin{aligned} a&=2,\quad b=3,\quad c=-4 &&\text{read signed coefficients}\\ x&=\frac{-3\pm\sqrt{3^2-4(2)(-4)}}{2(2)} &&\text{substitute}\\ &=\frac{-3\pm\sqrt{9+32}}4 &&\text{negative times negative is positive}\\ &=\frac{-3\pm\sqrt{41}}4 &&\text{simplify inside the root} \end{aligned}

Why keep $\sqrt{41}$ ? It is not an integer, and replacing it by a decimal would lose the requested exactness.

Answer. $x=\frac{-3+\sqrt{41}}4$ or $x=\frac{-3-\sqrt{41}}4$ .

Check. For either root, $4x+3=\pm\sqrt{41}$ , so $(4x+3)^2=41$ . Expanding gives $16x^2+24x-32=0$ , which divides by $8$ to give the original equation.

Change only the sign of the linear coefficient. Its square stays the same, but $-b$ changes sign.

Worked example 2: a negative linear coefficient

Question. Solve $2x^2-3x-4=0$ exactly.

Plan. Include $b=-3$ in both places where the formula uses $b$ .

\begin{aligned} a&=2,\quad b=-3,\quad c=-4 &&\text{keep the signs}\\ x&=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(-4)}}{2(2)} &&\text{substitute with brackets}\\ &=\frac{3\pm\sqrt{41}}4 &&\text{simplify both signs} \end{aligned}

Why the brackets? $(-3)^2=9$ , while $-3^2$ without brackets means $-9$ .

Answer. $x=\frac{3+\sqrt{41}}4$ or $x=\frac{3-\sqrt{41}}4$ .

Check. Now $(4x-3)^2=41$ , which expands to $8(2x^2-3x-4)=0$ .

Here is a useful surprise: the two irrational roots in Example 1 add to the rational number $-3/2$ , because their square-root terms cancel. This checks the paired signs without rounding either root.

Watch out

$x=-3\pm\frac{\sqrt{41}}4$ is not the answer to Example 1: only the root term is divided by $4$ . Keep the whole numerator in brackets: $x=(-3\pm\sqrt{41})/4$ .

Enter each whole numerator in calculator brackets. Show the formula substitution when algebraic working is requested.

Solving by completing the square (Higher)

Completing the square rewrites the quadratic as a squared bracket plus a constant. Solving then means isolating the square and taking both square roots.

Since $(x-3)^2=x^2-6x+9$ , the expression $x^2-6x+2$ equals $(x-3)^2-7$ : the $+9$ from the square must be reduced by $7$ to leave $+2$ .

Completing the square

x^2+bx+c=\left(x+\frac b2\right)^2+c-\left(\frac b2\right)^2

Half the linear coefficient goes inside the bracket. Subtract its square to preserve the original expression. A general quadratic's completed-square form is $a(x+b)^2+c$ , where the letters take the values needed for that expression.
Not in the Higher formulae sheet: learn the method.

If the $x^2$ coefficient is not $1$ , divide the whole equation by it first. Alternatively, factor it from the quadratic and linear terms. Do not divide only the squared term.

Worked example 3: taking both square roots

Question. Solve $x^2-6x+2=0$ by completing the square.

Plan. Use half of $-6$ inside the bracket, isolate the square and keep both signs.

\begin{aligned} (x-3)^2-9+2&=0 &&\text{add and subtract }3^2\\ (x-3)^2&=7 &&\text{collect constants}\\ x-3&=\pm\sqrt7 &&\text{both values square to }7\\ x&=3\pm\sqrt7 &&\text{add }3 \end{aligned}

Why both signs? The equation asks which numbers have square $7$ , and both $\sqrt7$ and $-\sqrt7$ do.

Answer. $x=3+\sqrt7$ or $x=3-\sqrt7$ .

Check. In either case $x-3=\pm\sqrt7$ , so $(x-3)^2-7=0$ ; expansion returns the original quadratic.

Change only the sign of the linear term. Half its coefficient becomes positive, so the bracket is $x+3$ and the final shift is negative.

Worked example 4: a positive number inside the bracket

Question. Solve $x^2+6x+2=0$ by completing the square.

Plan. Complete $(x+3)^2$ , then undo the square and the addition.

\begin{aligned} (x+3)^2-9+2&=0 &&\text{preserve the constant term}\\ (x+3)^2&=7 &&\text{isolate the square}\\ x+3&=\pm\sqrt7 &&\text{take both square roots}\\ x&=-3\pm\sqrt7 &&\text{subtract }3 \end{aligned}

Why subtract $3$ ? It reverses the $+3$ inside the bracket. The bracket number is not itself the value of $x$ .

Answer. $x=-3+\sqrt7$ or $x=-3-\sqrt7$ .

Check. Each root makes $(x+3)^2=7$ , hence $x^2+6x+2=0$ .

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2.7 - Quadratic formula and completing the square

Using the quadratic formula (Higher)

Solving by completing the square (Higher)

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