Year 11 Methods (Unit 1 & 2) Quadratics

The General Quadratic Formula

Q: What is the quadratic formula?

For a x squared plus b x plus c equals zero, x equals negative b plus or minus the square root of b squared minus four a c, all over two a. It gives both solutions of any quadratic equation.

Q: How do I get exact versus approximate solutions?

For exact solutions, simplify the square root of b squared minus four a c and leave it in surd form. For approximate solutions, work the square root out as a decimal and round each answer to the stated number of decimal places.

Q: Do I always get two solutions?

You get two solutions when b squared minus four a c is greater than zero, one repeated solution when it equals zero, and no real solutions when it is less than zero.

Q: How does the formula connect to the graph?

The solutions of a x squared plus b x plus c equals zero are the x-intercepts of the parabola y equals a x squared plus b x plus c.

Q: What is the first thing to do before substituting?

Rearrange the equation into standard form, a x squared plus b x plus c equals zero, and read off a, b and c with their correct signs.

20 practice questions 1 video lesson Theory + worked examples

Master the quadratic formula for Queensland Year 11 Mathematical Methods (QCAA). The quadratic formula solves any quadratic equation, even those that will not factorise neatly.

You will learn to rearrange into standard form, substitute the coefficients carefully, and give either exact surd solutions or rounded decimals — then read those solutions as the x-intercepts of the parabola, a reliable tool when factorising fails.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), the quadratic formula \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) solves any equation of the form \(ax^2+bx+c=0\). This page shows how to substitute \(a\), \(b\) and \(c\), give exact (surd) or approximate (decimal) solutions, rearrange to standard form first, and read the roots as the x-intercepts of the parabola.

A quadratic equation is any equation that can be written in the standard form \(ax^2+bx+c=0\), where \(a\neq 0\). The numbers \(a\), \(b\) and \(c\) are the coefficients: \(a\) multiplies \(x^2\), \(b\) multiplies \(x\), and \(c\) is the constant term.

The quadratic formula gives the solutions (the roots) directly from those coefficients. It works for every quadratic, even ones that do not factor over the integers, so it is the reliable fall-back method.

An exact solution is left in surd form, such as \(3+\sqrt{2}\); an approximate solution is a decimal rounded to a stated number of places.

Always reach standard form first. Move every term to one side so the equation reads \(ax^2+bx+c=0\), then read off \(a\), \(b\) and \(c\) — keeping their signs — before you substitute.

The two solutions of \(ax^2+bx+c=0\) are exactly the x-intercepts of \(y=ax^2+bx+c\).

The \(\pm\) places the roots symmetrically about the axis \(x=-\dfrac{b}{2a}\).

For \(ax^2+bx+c=0\) with \(a\neq0\):

\[x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\]

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

The quantity under the root sign is the discriminant:

\[\Delta=b^2-4ac\]

Δ = b^{2} - 4 a c

The \(\pm\) gives the two roots. Simplify the surd \(\sqrt{b^2-4ac}\) fully, then split into \(x=\dfrac{-b+\sqrt{\Delta}}{2a}\) and \(x=\dfrac{-b-\sqrt{\Delta}}{2a}\).

How to solve with the quadratic formula

Standardise: rearrange to \(ax^2+bx+c=0\) and read off \(a\), \(b\), \(c\) with their signs.
Substitute: put the values into \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) and evaluate \(b^2-4ac\) first.
Simplify: simplify the surd, then either leave the exact answer or round each root to the required number of decimal places.

Example 1 — Exact solutions

Solve \(x^2+5x+3=0\), giving exact solutions.

Solution

Standardise — read off the coefficients:

\(a\)	\(=\)	\(1\)
\(b\)	\(=\)	\(5\)
\(c\)	\(=\)	\(3\)

Substitute into the formula:

\(x\)	\(=\)	\(\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\)
	\(=\)	\(\dfrac{-5\pm\sqrt{5^2-4(1)(3)}}{2(1)}\)
	\(=\)	\(\dfrac{-5\pm\sqrt{25-12}}{2}\)
	\(=\)	\(\dfrac{-5\pm\sqrt{13}}{2}\)

Solutions: \(x=\dfrac{-5+\sqrt{13}}{2}\) or \(x=\dfrac{-5-\sqrt{13}}{2}\).

x = \frac{- 5 \pm \sqrt{13}}{2}

Example 2 — Approximate solutions

Solve \(2x^2-4x-3=0\), giving solutions correct to 2 decimal places.

Solution

Coefficients:

\(a\)	\(=\)	\(2\)
\(b\)	\(=\)	\(-4\)
\(c\)	\(=\)	\(-3\)

Substitute (mind the signs):

\(x\)	\(=\)	\(\dfrac{-(-4)\pm\sqrt{(-4)^2-4(2)(-3)}}{2(2)}\)
	\(=\)	\(\dfrac{4\pm\sqrt{16+24}}{4}\)
	\(=\)	\(\dfrac{4\pm\sqrt{40}}{4}\)

Evaluate each root, then round:

\(x\)	\(=\)	\(\dfrac{4+6.3246}{4}=2.58\)
\(x\)	\(=\)	\(\dfrac{4-6.3246}{4}=-0.58\)

Solutions: \(x\approx 2.58\) or \(x\approx -0.58\).

x \approx 2.58 or x \approx - 0.58

Example 3 — Rearrange first

Solve \(3x^2=2x+4\), giving exact solutions.

Solution

Standardise — move every term to the left:

\(3x^2-2x-4\)

\(=\)

\(0\)

So \(a=3\), \(b=-2\), \(c=-4\).

Substitute:

\(x\)	\(=\)	\(\dfrac{-(-2)\pm\sqrt{(-2)^2-4(3)(-4)}}{2(3)}\)
	\(=\)	\(\dfrac{2\pm\sqrt{4+48}}{6}\)
	\(=\)	\(\dfrac{2\pm\sqrt{52}}{6}\)

Simplify the surd \(\sqrt{52}=2\sqrt{13}\):

\(x\)	\(=\)	\(\dfrac{2\pm 2\sqrt{13}}{6}\)
	\(=\)	\(\dfrac{1\pm\sqrt{13}}{3}\)

Solutions: \(x=\dfrac{1+\sqrt{13}}{3}\) or \(x=\dfrac{1-\sqrt{13}}{3}\).

x = \frac{1 \pm \sqrt{13}}{3}

Example 4 — Clear a fraction, then solve

Solve \(x+2=\dfrac{5}{x}\), giving exact solutions.

Solution

Clear the fraction — multiply both sides by \(x\):

\(x(x+2)\)	\(=\)	\(5\)
\(x^2+2x\)	\(=\)	\(5\)

Standardise:

\(x^2+2x-5\)

\(=\)

\(0\)

So \(a=1\), \(b=2\), \(c=-5\).

Substitute and simplify \(\sqrt{24}=2\sqrt{6}\):

\(x\)	\(=\)	\(\dfrac{-2\pm\sqrt{2^2-4(1)(-5)}}{2(1)}\)
	\(=\)	\(\dfrac{-2\pm\sqrt{4+20}}{2}\)
	\(=\)	\(\dfrac{-2\pm 2\sqrt{6}}{2}\)
	\(=\)	\(-1\pm\sqrt{6}\)

Solutions: \(x=-1+\sqrt{6}\) or \(x=-1-\sqrt{6}\).

x = - 1 \pm \sqrt{6}

Common pitfalls

Losing a sign in \(-b\). If \(b=-4\), then \(-b=+4\). Write the substitution with brackets, \(-(-4)\), so the double negative is not dropped.

Forgetting to standardise. The formula only applies to \(ax^2+bx+c=0\). Move all terms to one side (and clear fractions) before reading off \(a,b,c\).

Rounding too early. For exact answers keep the surd; for decimals, carry extra figures through the working and round only the final root.

Only writing one root. The \(\pm\) produces two solutions — always give both unless the discriminant is zero.

Frequently asked questions

What is the quadratic formula?

For \(ax^2+bx+c=0\) it is \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\); it gives both solutions of any quadratic equation.

When should I use the quadratic formula instead of factorising?

Use it whenever the quadratic will not factorise neatly over the integers, or when you are asked for exact surd or decimal solutions.

How do I get exact versus approximate solutions?

For exact, simplify \(\sqrt{b^2-4ac}\) and leave the surd; for approximate, evaluate the root as a decimal and round each solution to the stated number of places.

Do I always get two solutions?

You get two when \(b^2-4ac>0\), one repeated solution when \(b^2-4ac=0\), and no real solutions when \(b^2-4ac<0\).

How does the formula connect to the graph?

The solutions of \(ax^2+bx+c=0\) are the \(x\)-intercepts of the parabola \(y=ax^2+bx+c\).

What is the first thing to do before substituting?

Rearrange the equation into standard form \(ax^2+bx+c=0\) and read off \(a\), \(b\) and \(c\) with their correct signs.

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