Quadratic functions and equations

A quadratic function has the form

$f(x)=ax^2+bx+c,\qquad a\ne 0.$

Its graph is a parabola. A quadratic equation asks when a quadratic expression has a specified value, most often when

$ax^2+bx+c=0.$

Quadratics connect algebra and graphs: solving $f(x)=0$ finds the $x$ coordinates where the graph $y=f(x)$ meets the $x$ axis.

Prerequisites

You should be able to expand and factorise algebraic expressions, solve GCSE quadratic equations, and interpret graphs of functions.

Reading a quadratic graph

For $y=ax^2+bx+c$:

• $a>0$ gives a minimum and a parabola opening upwards.
• $a<0$ gives a maximum and a parabola opening downwards.
• $c$ is the $y$ intercept because $y=c$ when $x=0$.
• The roots are the $x$ intercepts, if real roots exist.
• The axis of symmetry is

$x=-\frac{b}{2a}.$

The turning point lies on this line. Substitute $x=-b/(2a)$ into the function to find its $y$ coordinate.

Worked example: sketch from key features

Consider

$f(x)=x^2-6x+5.$

Factorising gives

$f(x)=(x-1)(x-5),$

so the roots are $1$ and $5$. The $y$ intercept is $f(0)=5$. The axis of symmetry is midway between the roots:

$x=\frac{1+5}{2}=3.$

At $x=3$,

$f(3)=9-18+5=-4.$

Therefore the turning point is $(3,-4)$. Since the coefficient of $x^2$ is positive, this is a minimum.

The completed form $f(x)=(x-3)^2-4$ reveals the same information immediately. See completing the square for the full method.

Choosing a solving method

There are three main exact methods.

1. Factorisation is usually fastest when integer or simple rational factors are visible.
2. Completing the square is especially useful when the turning point, range or graph is also needed.
3. The quadratic formula always works for a quadratic equation and is often best when factorisation is not obvious.

First rearrange the equation into the form $ax^2+bx+c=0$. A method applied before one side is zero will usually produce incorrect factors or coefficients.

Solving by factorisation

The zero product property says

$uv=0\quad\Longrightarrow\quad u=0\text{ or }v=0.$

Worked example: a monic quadratic

Solve

$x^2-7x+12=0.$

Find two numbers whose product is $12$ and whose sum is $-7$:

$x^2-7x+12=(x-3)(x-4).$

Hence

$(x-3)(x-4)=0,$

so

$x=3\quad\text{or}\quad x=4.$

Worked example: a non-monic quadratic

Solve

$6x^2+x-2=0.$

Since $6\times(-2)=-12$, split the middle term using numbers with product $-12$ and sum $1$:

$6x^2+4x-3x-2=0.$

Factorise by grouping:

$2x(3x+2)-(3x+2)=0,$ $(3x+2)(2x-1)=0.$

Therefore

$x=-\frac23\quad\text{or}\quad x=\frac12.$

The quadratic formula

For

$ax^2+bx+c=0,\qquad a\ne0,$

the solutions are

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

The whole of $-b\pm\sqrt{b^2-4ac}$ is divided by $2a$. Brackets are important when entering the formula into a calculator.

Worked example: two irrational roots

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

Here $a=3$, $b=4$ and $c=-2$:

$x=\frac{-4\pm\sqrt{4^2-4(3)(-2)}}{2(3)}$ $=\frac{-4\pm\sqrt{40}}6$ $=\frac{-4\pm2\sqrt{10}}6$ $=\frac{-2\pm\sqrt{10}}3.$

If decimal values are required, round only at the end. Premature rounding can make later calculations inaccurate.

Worked example: rearrange first

Solve $2x^2+7=9x$.

Write the equation with zero on one side:

$2x^2-9x+7=0.$

It now factorises:

$(2x-7)(x-1)=0.$

Thus $x=7/2$ or $x=1$. The formula would give the same result, but factorisation is shorter.

The discriminant

The expression inside the square root is the discriminant:

$\Delta=b^2-4ac.$

For a quadratic with real coefficients:

Condition	Roots	Graphical meaning
$\Delta>0$	Two distinct real roots	The parabola crosses the $x$ axis twice
$\Delta=0$	One repeated real root	The parabola touches the $x$ axis at its turning point
$\Delta<0$	No real roots	The parabola does not meet the $x$ axis

The discriminant describes the roots without requiring you to calculate them.

Worked example: classify the roots

For $2x^2-5x+4=0$,

$\Delta=(-5)^2-4(2)(4)=25-32=-7.$

Therefore the equation has no real roots. Since $a=2>0$, the whole parabola lies above the $x$ axis.

Worked example: find a parameter for a repeated root

Find the values of $k$ for which

$x^2+kx+9=0$

has a repeated root.

A repeated root requires $\Delta=0$:

$k^2-4(1)(9)=0,$ $k^2=36,$ $k=6\quad\text{or}\quad k=-6.$

Both signs matter. The corresponding quadratics are $(x+3)^2$ and $(x-3)^2$.

Root conditions and inequalities

Words in a question translate into precise discriminant conditions:

• two distinct real roots means $\Delta>0$;
• real roots means $\Delta\geq0$;
• equal roots or one repeated root means $\Delta=0$;
• no real roots means $\Delta<0$.

Worked example: a range of parameter values

Find the values of $p$ for which

$x^2+2px+p+3=0$

has two distinct real roots.

Here $a=1$, $b=2p$ and $c=p+3$. Require $\Delta>0$:

$(2p)^2-4(1)(p+3)>0,$ $4p^2-4p-12>0,$ $p^2-p-3>0.$

The boundary values solve $p^2-p-3=0$:

$p=\frac{1\pm\sqrt{13}}2.$

Because the quadratic in $p$ opens upwards, it is positive outside its roots. Hence

$\boxed{p<\frac{1-\sqrt{13}}2\quad\text{or}\quad p>\frac{1+\sqrt{13}}2}.$

Do not replace $>0$ with $\geq0$: equality would produce a repeated root, not two distinct roots.

Equations quadratic in another expression

An equation can have quadratic structure even when its highest power of $x$ is not $2$. Identify a repeated expression and substitute for it.

Worked example: a quartic in $x^2$

Solve

$x^4-5x^2+4=0.$

Let $u=x^2$. Then

$u^2-5u+4=0,$ $(u-1)(u-4)=0.$

So $u=1$ or $u=4$. Restore $u=x^2$:

$x^2=1\quad\text{or}\quad x^2=4.$

Therefore

$\boxed{x=-2,-1,1,2}.$

Stopping at $x^2=1$ and $x^2=4$ would leave the original equation unsolved.

Worked example: powers with a common base

Solve

$3^{2x}-10(3^x)+9=0.$

Let $u=3^x$. Since $3^{2x}=(3^x)^2$, the equation becomes

$u^2-10u+9=0,$ $(u-1)(u-9)=0.$

Thus $3^x=1$ or $3^x=9$, giving

$x=0\quad\text{or}\quad x=2.$

Always check restrictions on the substituted quantity. Here $u=3^x>0$, so a negative solution for $u$ would have to be rejected.

Quadratic models

Quadratic models occur when a quantity initially increases and later decreases, or when an area is formed from two changing linear dimensions. The algebraic roots may not all be meaningful in context.

Worked example: maximum area

A rectangle has perimeter $40$ m. If its length is $x$ m, its width is $(20-x)$ m, so its area is

$A(x)=x(20-x)=-x^2+20x.$

Complete the square:

$A(x)=-(x-10)^2+100.$

Since $(x-10)^2\geq0$, the maximum area is

$\boxed{100\text{ m}^2},$

attained when $x=10$. The domain is $0<x<20$, because both side lengths must be positive.

Worked example: interpreting roots

The height of a ball above the ground is modelled by

$h(t)=-5t^2+20t+1,$

where $t$ is time in seconds. To find when it reaches the ground, solve $h(t)=0$:

$t=\frac{-20\pm\sqrt{20^2-4(-5)(1)}}{2(-5)}$ $=\frac{-20\pm\sqrt{420}}{-10}.$

This gives approximately $t=-0.05$ or $t=4.05$. The negative value is algebraically valid but outside the model’s domain $t\geq0$. The ball reaches the ground after approximately $4.05$ seconds.

For more on domains, assumptions and interpreting parameters, see functions in modelling.

Connections between roots and coefficients

If the roots of $ax^2+bx+c=0$ are $\alpha$ and $\beta$, then

$a(x-\alpha)(x-\beta)=ax^2-a(\alpha+\beta)x+a\alpha\beta.$

Comparing coefficients gives

$\alpha+\beta=-\frac ba,\qquad \alpha\beta=\frac ca.$

These relations can answer questions about roots without solving the equation.

Worked example: build a quadratic from its roots

Find a monic quadratic whose roots are $2+\sqrt3$ and $2-\sqrt3$.

Their sum and product are

$\alpha+\beta=4,$ $\alpha\beta=(2+\sqrt3)(2-\sqrt3)=1.$

Therefore the quadratic is

$\boxed{x^2-4x+1=0}.$

Common misconceptions

• $ax^2+bx+c$ is quadratic only when $a\ne0$.
• Factorisation solves an equation only after it has been set equal to zero.
• From $(x-r)(x-s)=0$, the roots are $r$ and $s$, not $-r$ and $-s$.
• In the formula, use the signed coefficient. For $3x^2-5x+1$, $b=-5$.
• $\Delta=0$ gives one repeated root, not two distinct roots.
• The axis of symmetry is $x=-b/(2a)$, not a root in general.
• A substituted variable must be converted back to $x$, and restrictions must be checked.
• A mathematically valid root may be impossible in a model because of time, length or domain restrictions.

Check your understanding

1. Solve $2x^2-3x-5=0$.
2. Find the turning point and axis of symmetry of $y=2x^2+8x+3$.
3. Determine the number of real roots of $4x^2+4x+3=0$.
4. Find the values of $k$ for which $x^2+(k-1)x+k=0$ has a repeated root.
5. Solve $x^4-13x^2+36=0$.
6. A rectangle has area $48$ cm$^2$ and its length is $2$ cm greater than its width. Find its dimensions.

Answers

1. $(2x-5)(x+1)=0$, so $x=5/2$ or $x=-1$.
2. $2x^2+8x+3=2(x+2)^2-5$, so the turning point is $(-2,-5)$ and the axis is $x=-2$.
3. $\Delta=4^2-4(4)(3)=-32$, so there are no real roots.
4. $\Delta=(k-1)^2-4k=k^2-6k+1=0$, so $k=3\pm2\sqrt2$.
5. Let $u=x^2$. Then $(u-4)(u-9)=0$, so $x=-3,-2,2,3$.
6. If the width is $x$, then $x(x+2)=48$. Thus $(x+8)(x-6)=0$. Rejecting the negative length gives width $6$ cm and length $8$ cm.

What to learn next

• Study completing the square in detail.
• Learn how quadratics behave under graph transformations.
• Apply quadratic sign diagrams when solving inequalities.
• Extend factor and root ideas to polynomials.