Solving Quadratic Equations Using The Quadratic Formula
Every question with a fully worked solution.
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Theory
The quadratic formula solves any quadratic equation \(ax^2+bx+c=0\), even one that does not factorise: \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\). This Year 10 maths page covers exact surd answers, decimal answers, the discriminant \(b^2-4ac\), equations that need rearranging first and quadratic models where a solution is rejected.
Every quadratic equation can be written in standard form \(ax^2+bx+c=0\), where \(a\neq 0\). The coefficients \(a\), \(b\) and \(c\) include their signs: in \(2x^2-7x-3=0\), \(a=2\), \(b=-7\) and \(c=-3\).
The quadratic formula gives both solutions at once. The \(\pm\) sign means one solution uses \(+\) and the other uses \(-\).
The expression under the square root, \(\Delta=b^2-4ac\), is the discriminant. If \(\Delta\gt 0\) there are two real solutions, if \(\Delta=0\) there is one, and if \(\Delta\lt 0\) there are none.
An exact answer keeps the surd, such as \(4\pm\sqrt{5}\). A decimal approximation rounds each solution separately, such as \(6.24\) and \(1.76\). If \(\Delta\) is a perfect square, the solutions are rational and the quadratic could also have been factorised.
For \(ax^2+bx+c=0\) with \(a\neq 0\):
| Discriminant | Number of real solutions | Graph of \(y=ax^2+bx+c\) |
|---|---|---|
| \(\Delta\gt 0\) | two (different) | crosses the \(x\)-axis twice |
| \(\Delta=0\) | one (repeated) | touches the \(x\)-axis |
| \(\Delta\lt 0\) | none | does not meet the \(x\)-axis |
How to use the quadratic formula
- Rearrange into standard form \(ax^2+bx+c=0\). Expand brackets, clear fractions (state the excluded values) and clear decimals.
- Write down \(a\), \(b\) and \(c\) with their signs.
- Work out \(\Delta=b^2-4ac\) on its own line. Use brackets for a negative \(b\): \((-6)^2=36\). If \(\Delta\lt 0\), stop: there are no real solutions.
- Substitute into \(x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\).
- Simplify the surd and cancel any common factor for an exact answer, or use a calculator for each sign separately for a decimal answer.
- Check the context: reject a negative length or time, and any value that makes a denominator zero.
Setting up word problems
- Perimeter and area: if the perimeter is \(P\), the two sides add to \(\dfrac{P}{2}\), so they are \(x\) and \(\dfrac{P}{2}-x\).
- Folded boxes: cutting a square of side \(x\) from each corner of an \(L\) by \(W\) sheet leaves a base \((L-2x)\) by \((W-2x)\) and a height \(x\). Both base lengths must be positive.
- Pythagoras: with sides written in \(x\), use \(a^2+b^2=c^2\) with the hypotenuse on its own side, then simplify to standard form.
- Height models: set \(h\) equal to the height asked for. Use the unrounded solution in any later calculation.
(i) \(2x^2+5x-1=0\)
(ii) \(x^2-8x+11=0\)
(i) \(a=2\), \(b=5\), \(c=-1\).
| \(x\) | \(=\) | \(\dfrac{-5\pm\sqrt{5^2-4\times 2\times(-1)}}{2\times 2}\) |
| \(\) | \(=\) | \(\dfrac{-5\pm\sqrt{25+8}}{4}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{-5\pm\sqrt{33}}{4}}\) |
(ii) \(a=1\), \(b=-8\), \(c=11\). Then \(-b=8\).
| \(x\) | \(=\) | \(\dfrac{8\pm\sqrt{64-44}}{2}\) |
| \(\) | \(=\) | \(\dfrac{8\pm\sqrt{20}}{2}\) |
| \(\) | \(=\) | \(\dfrac{8\pm 2\sqrt{5}}{2}\) |
| \(\) | \(=\) | \(\boldsymbol{4\pm\sqrt{5}}\) |
Simplify the surd first (\(\sqrt{20}=2\sqrt{5}\)), then divide every term of the numerator by \(2\).
(i) \((x-2)(x+3)=4x\), exactly and to \(2\) decimal places
(ii) \(2x-\dfrac{1}{x}=3\), exactly
(i) Expand and make one side zero.
| \(x^2+x-6\) | \(=\) | \(4x\) |
| \(x^2-3x-6\) | \(=\) | \(0\) |
\(a=1\), \(b=-3\), \(c=-6\):
| \(x\) | \(=\) | \(\dfrac{3\pm\sqrt{9+24}}{2}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{3\pm\sqrt{33}}{2}}\) |
| \(\) | \(\approx\) | \(\boldsymbol{4.37} \text{ or } \boldsymbol{-1.37}\) |
(ii) \(x\neq 0\). Multiply every term by \(x\), then make one side zero.
| \(2x^2-1\) | \(=\) | \(3x\) |
| \(2x^2-3x-1\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\dfrac{3\pm\sqrt{9+8}}{4}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{3\pm\sqrt{17}}{4}}\) |
Neither value is \(0\), so both are solutions.
(i) \(3x^2-4x+2=0\)
(ii) \(9x^2-30x+25=0\)
(i) \(\Delta\) \(= b^2-4ac\) \(= (-4)^2-4\times 3\times 2\) \(= 16-24\) \(= -8\).
\(\Delta\lt 0\). A negative number has no real square root, so there are no real solutions.
(ii) \(\Delta\) \(= (-30)^2-4\times 9\times 25\) \(= 900-900\) \(= 0\).
| \(x\) | \(=\) | \(\dfrac{30\pm\sqrt{0}}{18}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{5}{3}}\) |
\(\Delta=0\), so there is exactly one solution. Check: \(9x^2-30x+25=(3x-5)^2\).
(i) when does it hit the ground?
(ii) when is it \(6\) m above the ground?
(i) Set \(h=0\) and write the equation with a positive \(t^2\) term: \(4.9t^2-12t-1.5=0\).
| \(t\) | \(=\) | \(\dfrac{12\pm\sqrt{144+4\times 4.9\times 1.5}}{9.8}\) |
| \(\) | \(=\) | \(\dfrac{12\pm\sqrt{173.4}}{9.8}\) |
| \(t\) | \(\approx\) | \(2.568 \text{ or } -0.119\) |
Time after the throw cannot be negative, so reject \(-0.119\). The ball lands after \(\boldsymbol{2.57}\) seconds.
(ii) Set \(h=6\): \(4.9t^2-12t+4.5=0\).
| \(t\) | \(=\) | \(\dfrac{12\pm\sqrt{144-88.2}}{9.8}\) |
| \(\) | \(=\) | \(\dfrac{12\pm\sqrt{55.8}}{9.8}\) |
| \(t\) | \(\approx\) | \(\boldsymbol{0.46} \text{ s or } \boldsymbol{1.99}\text{ s}\) |
Both are valid: the ball passes \(6\) m on the way up and again on the way down.
Common pitfalls
Frequently asked questions
What is the quadratic formula?
For \(ax^2+bx+c=0\), \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\). It solves every quadratic equation, including ones that do not factorise.
What does the discriminant tell you?
The discriminant \(b^2-4ac\) tells you the number of real solutions: two if it is positive, one if it is zero, and none if it is negative.
When should I use the formula instead of factorising?
Factorise when the numbers work out easily. Use the formula when the quadratic does not factorise, when the question asks for surds or decimals, or when you cannot see the factors quickly.
How do I simplify an answer like (6 plus or minus 2 root 5) over 2?
Every term in the numerator has a factor of \(2\), so divide each one by \(2\) to get \(3\pm\sqrt{5}\). Simplify the surd before cancelling.
Why is one solution rejected in word problems?
The formula gives every solution of the equation, but the situation may not allow some of them. A length or a time after the start cannot be negative, and a box cannot have a side of negative length.