Solving quadratic equations, including disguised quadratics

What you need to be able to do

What you should already know

This page builds on four O Level skills. Do the quick check on each card; if one feels shaky, fix it first, because every method below uses it.

What a quadratic equation is

\(a\), \(b\) and \(c\) are the coefficients; read them with their signs. In \(3x^2-x+5=0\): \(a=3,\ b=-1,\ c=5\). In \(4-x^2=0\): \(a=-1,\ b=0,\ c=4\). If an equation isn't in this shape yet (\(x^2=3x+10\), \(x(x+2)=8\)), rearrange it first: everything on one side, \(0\) on the other.

Why we make one side zero: the zero-product rule

So once \(x^2-7x+12=0\) is written as \((x-3)(x-4)=0\), one bracket must be zero: \(x=3\) or \(x=4\). The rule does not work for any other number. \((x-3)(x-4)=2\) does not mean \(x-3=2\) or \(x-4=2\) (try \(x=5\): \(2\times1=2\) ✓, but \(x=6\): \(3\times2=6\) ✗). Expand and rearrange first: \(x^2-7x+10=0\), so \(x=2\) or \(5\).

Method 1: factorising

Factorising is the quickest method when it works, and it works whenever the roots are whole numbers or fractions. There are four shapes to recognise.

A repeated root. \(x^2-6x+9=(x-3)^2=0\) gives \(x=3\) twice. The graph touches the \(x\)-axis at one point instead of crossing it. You will meet this again with the discriminant.

Method 2: completing the square

Every quadratic can be written as a perfect square plus a number (the page Completing the square teaches it in full). Once it is in that form, you solve it by isolating the square and taking a square root with \(\pm\).

1. Solve \(x^2+6x-4=0\). Half of \(6\) is \(3\): \(x^2+6x=(x+3)^2-9\).
2. So \((x+3)^2-9-4=0\Rightarrow(x+3)^2=13\)
3. Square-root both sides, with \(\pm\): \(x+3=\pm\sqrt{13}\)
4. \(x=-3+\sqrt{13}\) or \(x=-3-\sqrt{13}\)

Completing the square also shows at a glance when there are no real roots. \(x^2+4x+7=(x+2)^2+3\). Setting it to zero needs \((x+2)^2=-3\), and a square is never negative, so there are no real roots.

Method 3: the quadratic formula, and where it comes from

Completing the square on the general equation \(ax^2+bx+c=0\) gives a formula that solves every quadratic. Follow the derivation once; it shows why the formula works and where the \(\pm\) comes from.

1. Divide by \(a\): \(x^2+\dfrac bax+\dfrac ca=0\)
2. Complete the square (half of \(\tfrac ba\) is \(\tfrac b{2a}\)): \(\left(x+\dfrac b{2a}\right)^2-\dfrac{b^2}{4a^2}+\dfrac ca=0\)
3. Isolate the square, over a common denominator: \(\left(x+\dfrac b{2a}\right)^2=\dfrac{b^2-4ac}{4a^2}\)
4. Square-root with \(\pm\): \(x+\dfrac b{2a}=\pm\dfrac{\sqrt{b^2-4ac}}{2a}\)
5. So \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Using it. Solve \(3x^2-4x-2=0\): \(a=3,\ b=-4,\ c=-2\).

1. \(x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(3)(-2)}}{2(3)}\) (brackets round every negative)
2. \(=\dfrac{4\pm\sqrt{16+24}}{6}=\dfrac{4\pm\sqrt{40}}{6}\)
3. \(\sqrt{40}=2\sqrt{10}\), so \(x=\dfrac{4\pm2\sqrt{10}}{6}=\dfrac{2\pm\sqrt{10}}{3}\)
4. As decimals (only if asked): \(x\approx1.72\) or \(x\approx-0.387\)

Which method should I use?

Equations that turn into quadratics

An equation with \(x\) in a denominator is often a quadratic in disguise. Multiply every term by the denominator(s) to clear the fractions, rearrange to \(=0\), then solve. Finally check that no answer makes a denominator zero.

Disguised quadratics: substitute, solve, then undo

An equation is a disguised quadratic when one power is the square of another: \(x^4\) and \(x^2\); \(x^6\) and \(x^3\); \(x\) and \(\sqrt x\); \(u^2\) and \(u\) for any expression \(u\). Replace the smaller one with a new letter and it becomes an ordinary quadratic.

For example \(x^4-13x^2+36=0\). Since \(x^4=(x^2)^2\), let \(u=x^2\): \(u^2-13u+36=0\Rightarrow(u-4)(u-9)=0\), so \(u=4\) or \(9\). But \(u\) was \(x^2\): \(x^2=4\Rightarrow x=\pm2\) and \(x^2=9\Rightarrow x=\pm3\). Four solutions: \(-3,-2,2,3\).

1. Spot the pair: which expression, squared, gives the other? Call it \(u\) (or \(y\), \(k\): not \(x\)).
2. Rewrite the equation in \(u\) and rearrange to \(au^2+bu+c=0\). Multiply out any fractions first.
3. Solve for \(u\) with any of the three methods.
4. Undo the substitution: each value of \(u\) gives an equation in \(x\) (\(x^2=5\), \(x^3=-27\), \(\sqrt x=\tfrac12\), …). Solve each one.
5. Reject impossible values and say why: \(x^2=-1\) and \(\sqrt x=-2\) have no real solutions. Keep negative values that are possible: \(x^3=-27\) gives \(x=-3\).

Key facts at a glance

• A quadratic equation \(ax^2+bx+c=0\) has at most two real roots: where \(y=ax^2+bx+c\) meets the \(x\)-axis.
• Get \(0\) on one side before factorising (zero-product rule). Never divide by \(x\).
• Three methods: factorise, complete the square, formula. The formula is completing the square done once in general.
• \(b^2-4ac<0\) or \((x+p)^2=\) negative: no real roots.
• Fractions: multiply every term by the denominators. Disguised: substitute, solve, undo, reject the impossible.

The rule that decides the marks: show your method

Every examiner report since 2020 repeats this. 9709/11/M/J/24 Q1(b): "Use of calculator with no working scores 0/3." Factors must also be real factors: the June 2021 report saw candidates get the roots from a calculator and then write "factors" that did not expand back to the quadratic, and lost the method mark.

"Hence" after a completed square is a hint to use it: the June 2022 report noted that most candidates ignored "hence" and used the formula. It still scored there, but on harder questions ignoring "hence" costs marks. A fraction equation is usually an opening 3-mark question; the November 2022 report saw candidates fail to multiply out fully, or add a number instead of subtracting it.

Mark-scheme focus

Solve … (clear the fraction, then a quadratic in x³)

Credited

• multiply by \(x^3\) and recognise a quadratic in \(x^3\): \(u^2-28u+27=0\) M1*
• a seen method to solve it: \((x^3-1)(x^3-27)=0\) DM1
• \(x=1,\ 3\) B1

Not credited

• \(x^3=1,\ 27\) given as the answer (the substitution not undone)
• \(x=\pm1,\ \pm3\): a cube root has one real value
• solutions from a calculator with no factorising or formula

Why

Two steps students skip: clear the fraction first (×\(x^3\)), and at the end take the cube root of BOTH values of \(u\).

From: 9709/15/O/N/25 Q5.

Solve … with a square root: undo the substitution

Credited

• \((4x^{\frac12}-3)(x^{\frac12}-2)=0\), or \(u=x^{\frac12}\) and \(4u^2-11u+6=0\) solved M1
• \(x^{\frac12}=\tfrac34\) or \(2\) A1
• squared: \(x=\tfrac9{16}\) or \(4\) A1

Not credited

• \(x=\tfrac34,\ 2\) (the values of \(\sqrt x\), not \(x\)): M1 A0 A0
• spotting one solution by trial
• "showing all necessary working" ignored

Why

The letter you solved for is \(\sqrt x\). The question asks for \(x\), so square each value. Both are positive, so both are kept.

From: 9709/11/O/N/18 Q1.

Worked examples

Example 1 · Completed square, then a quartic (Paper 1, 5 marks) ★★ Medium

1 Read it with me what is actually happening?

Part (a) is a normal completed square, but in the letter \(y\). Part (b) has \(x^4\) and \(x^2\): it is the same quadratic with \(y=x^2\). So (a) already did most of the work.

• Henceuse (a): the quartic is (a) with \(y\) replaced by \(x^2\)
• exactanswers as surds: \(\pm\sqrt5\)
• \(x^4\)\(x^4=(x^2)^2\): a quadratic in \(x^2\)

2 What is it asking for? the target, part by part

• (a)\(a\) and \(b\)2 marks
• (b)all real \(x\)3 marks

3 What do we know? everything the question gives us

• \(3y^2-12y-15\)(a)
• \(3x^4-12x^2-15=0\)(b)

• \(x^2\) can never be negative: any negative value of \(x^2\) gives no real \(x\)

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• (a)\(3(y+a)^2+b\)
  • Use\(3(y^2-4y)-15\), half of −4 is −2
• (b)exact \(x\)
  • Use\(y=x^2\) in (a): \(3(x^2-2)^2-27=0\)
    • Then\((x^2-2)^2=9\Rightarrow x^2-2=\pm3\)
  • Trap\(x^2=5\) gives TWO values, \(\pm\sqrt5\)

Why is x² = −1 thrown away?

No real number squares to a negative. It gives no real solutions, so the only answers come from \(x^2=5\).

How many real solutions does the quartic have?

Two: \(\sqrt5\) and \(-\sqrt5\).

5 Now solve it have a go on paper first, then open the worked solution

1. (a) \(3(y^2-4y)-15=3[(y-2)^2-4]-15=\mathbf{3(y-2)^2-27}\): \(a=-2,\ b=-27\) B1 B1
2. (b) With \(y=x^2\): \(3(x^2-2)^2-27=0\Rightarrow(x^2-2)^2=9\Rightarrow x^2-2=\pm3\) M1
3. \(x^2=5\) or \(x^2=-1\) (no real solutions) M1
4. \(x=\pm\sqrt5\) A1

6 Check it does the answer make sense?

\(x^2=5\): \(3(25)-12(5)-15=75-60-15=0\) ✓.

Example 2 · A sixth power and a cube (Paper 1, 3 marks) ★★ Medium

1 Read it with me what is actually happening?

\(x^6=(x^3)^2\), so this is a quadratic in \(x^3\). Solve for \(x^3\), then take cube roots.

• \(x^6\) and \(x^3\)one is the square of the other: let \(u=x^3\)
• Solveshow the factorisation or the formula: 3 marks for working

2 What is it asking for? the target, part by part

• every real \(x\)3 marks

3 What do we know? everything the question gives us

• \(8x^6+215x^3-27=0\)

• cube roots of negative numbers exist: \(\sqrt[3]{-27}=-3\)

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• Findx
  • Use\(u=x^3\): \(8u^2+215u-27=0\)
    • Factorise\(8\times(-27)=-216\): \(216\) and \(-1\)
  • Trapthrowing away \(x^3=-27\)

Which two numbers multiply to −216 and add to 215?

216 and −1, so \(8u^2+216u-u-27=8u(u+27)-(u+27)=(8u-1)(u+27)\).

Is \(\sqrt[3]{1/8}=\pm\tfrac12\)?

No. Cube roots have one real value: \(\tfrac12\). Only even roots come with ±.

5 Now solve it have a go on paper first, then open the worked solution

1. Let \(u=x^3\): \(8u^2+215u-27=0\Rightarrow(8u-1)(u+27)=0\) M1
2. \(u=\tfrac18\) or \(u=-27\) A1
3. \(x^3=\tfrac18\Rightarrow x=\tfrac12\); \(x^3=-27\Rightarrow x=-3\) A1

6 Check it does the answer make sense?

\(x=-3\): \(8(729)+215(-27)-27=5832-5805-27=0\) ✓.

Example 3 · Substitution for a whole bracket (Paper 1, 4 marks) ★★ Medium

1 Read it with me what is actually happening?

The bracket \((2x-3)^2\) appears twice, once on top of a fraction. Calling it \(u\) turns the equation into \(u-\tfrac4u-3=0\), which becomes a quadratic once you multiply by \(u\).

• suitable substitutionlet \(u=(2x-3)^2\) (or \(u=2x-3\), then it is a quartic in \(u\))
• 4/(2x − 3)²multiply every term by \(u\) to clear it

2 What is it asking for? the target, part by part

• all values of \(x\)4 marks

3 What do we know? everything the question gives us

• \((2x-3)^2-\dfrac{4}{(2x-3)^2}-3=0\)

• \(u=(2x-3)^2\) is a square, so it can't be negative

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• Findx
  • Use\(u^2-3u-4=0\)
    • Then\(u=4\) or \(u=-1\)
  • Undo\((2x-3)^2=4\Rightarrow2x-3=\pm2\)
  • Trap\((2x-3)^2=-1\) has no real solution

What do you get when you multiply u − 4/u − 3 = 0 by u?

\(u^2-4-3u=0\), i.e. \(u^2-3u-4=0\).

Why are there two answers for x from u = 4?

\(2x-3\) can be \(+2\) or \(-2\).

5 Now solve it have a go on paper first, then open the worked solution

1. \(u=(2x-3)^2\): \(u^2-3u-4=0\) M1
2. \((u-4)(u+1)=0\Rightarrow u=4\) (\(u=-1\) rejected) M1
3. \(2x-3=\pm2\) A1
4. \(x=\tfrac52\) or \(x=\tfrac12\) only A1

6 Check it does the answer make sense?

\(x=\tfrac52\): \((2)^2-\tfrac44-3=4-1-3=0\) ✓. \(x=\tfrac12\): \((-2)^2-\tfrac44-3=0\) ✓.

Example 4 · A square root and its reciprocal (Paper 1, 4 marks) ★★ Medium

1 Read it with me what is actually happening?

\(\sqrt y\) and \(\tfrac{1}{\sqrt y}\): multiply every term by \(\sqrt y\) and you get \(y\), a constant and \(\sqrt y\). Since \(y=(\sqrt y)^2\), that is a quadratic in \(\sqrt y\).

• \(\tfrac{2}{\sqrt y}\)multiply through by \(\sqrt y\): \(\tfrac{2}{\sqrt y}\times\sqrt y=2\)
• Solve4 marks: two for the values of √y, two for y

2 What is it asking for? the target, part by part

• (a)both values of \(y\)4 marks

3 What do we know? everything the question gives us

• \(6\sqrt y+\dfrac2{\sqrt y}-7=0\)

• \(\sqrt y\) must be positive (both roots here are)

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• Findy
  • Use× \(\sqrt y\): \(6y-7\sqrt y+2=0\)
    • Let\(u=\sqrt y\): \(6u^2-7u+2=0\)
  • Undosquare each \(u\)

6 × 2 = 12: which two numbers multiply to 12 and add to −7?

\(-3\) and \(-4\): \(6u^2-3u-4u+2=3u(2u-1)-2(2u-1)=(3u-2)(2u-1)\).

If \(\sqrt y=\tfrac23\), what is \(y\)?

\(\tfrac49\): square both sides.

5 Now solve it have a go on paper first, then open the worked solution

1. \(\times\sqrt y\): \(6y+2-7\sqrt y=0\), a 3-term quadratic in \(\sqrt y\) M1
2. \((2\sqrt y-1)(3\sqrt y-2)=0\) DM1
3. \(\sqrt y=\tfrac12\) or \(\tfrac23\) A1
4. \(y=\tfrac14\) or \(y=\tfrac49\) A1

Part (b) then solves \(6\sqrt{\tan x}+\tfrac{2}{\sqrt{\tan x}}-7=0\): the same equation with \(y=\tan x\), so \(\tan x=\tfrac14\) or \(\tfrac49\) (see Trigonometry).

6 Check it does the answer make sense?

\(y=\tfrac14\): \(6(\tfrac12)+\tfrac{2}{1/2}-7=3+4-7=0\) ✓.

Example 5 · Surd answers from a square-root equation (Paper 1, 4 marks) ★★★ Hard

Why it is hard: the March 2022 report: many found only one value, and the answer must come from the formula or completing the square, not a calculator.

1 Read it with me what is actually happening?

\(x-4x^{1/2}+1=0\) is a quadratic in \(\sqrt x\). It doesn't factorise, so the formula (or completing the square) gives \(\sqrt x\) as a surd; squaring that gives \(x\).

• \(x^{1/2}\)\(\sqrt x\); and \(x=(\sqrt x)^2\)
• \(a+b\sqrt c\)integers: you must square the surd and simplify

2 What is it asking for? the target, part by part

• (a)both values of \(x\)4 marks

3 What do we know? everything the question gives us

• \(x-4\sqrt x+1=0\)\(x\ge0\)

• both values of \(\sqrt x\) must be checked: \(\sqrt x\) cannot be negative

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• Find\(x=a+b\sqrt c\)
  • Use\(u=\sqrt x\): \(u^2-4u+1=0\)
    • Formula\(u=\tfrac{4\pm\sqrt{12}}{2}=2\pm\sqrt3\)
  • Undo\(x=(2\pm\sqrt3)^2\)
  • Trapdiscarding \(2-\sqrt3\): it is positive (0.27)

Expand \((2+\sqrt3)^2\).

\(4+4\sqrt3+3=7+4\sqrt3\).

Is \(2-\sqrt3\) an allowed value of \(\sqrt x\)?

Yes: \(\sqrt3\approx1.73\), so \(2-\sqrt3>0\). Both roots are valid.

5 Now solve it have a go on paper first, then open the worked solution

1. \(u=\sqrt x\): \(u^2-4u+1=0\Rightarrow u=\dfrac{4\pm\sqrt{16-4}}{2}=2\pm\sqrt3\) M1 A1
2. \(x=(2\pm\sqrt3)^2\) M1
3. \(x=7+4\sqrt3\) or \(x=7-4\sqrt3\) A1

Alternative on the scheme: isolate the root, \(x+1=4\sqrt x\), square: \(x^2+2x+1=16x\Rightarrow x^2-14x+1=0\), \(x=7\pm4\sqrt3\). (Squaring can create false roots, so check both: here both work.)

6 Check it does the answer make sense?

\(7-4\sqrt3\approx0.072\): \(0.072-4(0.268)+1\approx0\) ✓.

Summary

• Solve by factorising, the formula (substitution shown) or completing the square; never by calculator alone.
• Rearrange to = 0 first; clear fractions by multiplying through.
• Exact = surds and fractions; ± gives two answers.
• Disguised quadratic: let \(u\) = the smaller power, solve for \(u\), undo, reject impossible values (\(x^2<0\), \(\sqrt x<0\)).

Can I do this?

Quiz

Concept checks

Answer each one in your head first, then open it.

Exam-style practice

New questions, not used above. Do them on paper, then mark your work against the official mark scheme.