Do I really have to show working if I can see the answer?

Yes, because a bare answer usually scores only the answer mark. Write one operation per line: the factorisation, the two equations x + 7 = 0 and x - 2 = 0, and the substitution check at the end.

What if b2 - 4ac comes out negative?

Then there are no real roots and you can stop. x2 + x + 1 has 1 - 4 = -3, so its parabola never crosses the x-axis, and writing 'no real solutions' is the complete answer.

Can a quadratic have three roots?

No. A quadratic has exactly two roots in total, so the only possibilities are two real roots, one repeated real root or none.

Which method should I use first in an exam?

Try factorisation first, because it is fastest when b2 - 4ac is a perfect square like 81. If no factor pair appears within about a minute, switch to the formula, which always works.

Solve it, show the working

Xai60p — Step-by-step solving of quadratic and | xai60p.com

Take x2 + 5x - 14 = 0. You need two numbers that multiply to -14 and add to 5: those are 7 and -2, so the equation factors as (x + 7)(x - 2) = 0 and the roots are x = -7 and x = 2. Below, the same equation is solved three ways, with the check lines an examiner expects at the end.

Solve

Factorisation, the formula and completing the square, each worked on a real equation from the first line to the checked answer.

Decide

The discriminant b2 - 4ac sorts every quadratic into two roots, one repeated root or no real roots before you commit to a method.

Sketch and pair

Turning points and intercepts for y = ax2 + bx + c, then elimination for simultaneous pairs such as x + y = 7 and x - y = 1.

A quadratic has exactly two roots in total: two, one or none of them real, but never three. One test, b2 - 4ac, decides which case you are in.

The check is part of the answer: 49 - 35 - 14 = 0 is the line that proves x = -7.

The worked answer: x2 + 5x - 14 = 0 by factorisation

Find two numbers that multiply to -14 and add to 5, split the equation into two brackets, then check both roots.

The method hangs on one observation: if (x + p)(x + q) = 0, then p x q = c and p + q = b. Here c = -14 and b = 5, so you need a pair that multiplies to -14 and adds to 5. The factor pairs of 14 are 1 x 14 and 2 x 7, and only 7 and 2 differ by 5; with the larger number positive, 7 and -2 add to +5.

That gives (x + 7)(x - 2) = 0. A product is zero only when one of its factors is zero, so the equation splits into x + 7 = 0 or x - 2 = 0, and the roots are x = -7 and x = 2.

The final line is the check, and it earns its own mark: 49 - 35 - 14 = 0 for x = -7, and 4 + 10 - 14 = 0 for x = 2. Both roots satisfy the original equation, so the solution is complete.

Step 1, about 30 seconds: write a = 1, b = 5, c = -14; typical failure is the sign slip, copying c as 14.
Step 2, about 1 minute: list the factor pairs of 14, 1 x 14 and 2 x 7; typical failure is stopping before testing which pair combines to +5.
Step 3, about 1 minute: choose 7 and -2 and write (x + 7)(x - 2) = 0; typical failure is the reversed bracket (x - 7)(x + 2), which gives -5x.
Step 4, about 30 seconds: set each bracket to zero, x = -7 or x = 2; typical failure is dropping the minus and writing x = 7.
Step 5, about 1 minute: check 49 - 35 - 14 = 0 and 4 + 10 - 14 = 0; typical failure is skipping the check and losing the final mark.

Six worked cards, one per skill

Each card is a complete mini-solution whose layout can be copied directly into an exercise book.

The key terms of this guide, drawn to one scale

The routines, side by side

The routines compared on one scale: lines of working, typical time and the check that closes each.
Routine	Lines of working	Typical time	Final check	Works best when
Factorisation	5 lines	about 4 minutes	substitute both roots back	b2 - 4ac is a perfect square, e.g. 81
Quadratic formula	6 lines	about 5 minutes	substitute both roots back	always, even with decimal roots
Completing the square	6 lines	about 5 minutes	expand the bracket back	roots involve a surd, e.g. sqrt(14)
Discriminant test	2 lines	about 1 minute	compare the value with 0	only the number of roots is needed
Elimination of a pair	4 lines	about 3 minutes	substitute the pair into both equations	one variable has matching coefficients
Vertex form	4 lines	about 3 minutes	expand to match the original	the turning point is needed, e.g. (2, -2)

The routines compared on one scale: lines of working, typical time and the check that closes each.

Asked and answered

Do I really have to show working if I can see the answer?: Yes, because a bare answer usually scores only the answer mark. Write one operation per line: the factorisation, the two equations x + 7 = 0 and x - 2 = 0, and the substitution check at the end.
What if b2 - 4ac comes out negative?: Then there are no real roots and you can stop. x2 + x + 1 has 1 - 4 = -3, so its parabola never crosses the x-axis, and writing 'no real solutions' is the complete answer.
Can a quadratic have three roots?: No. A quadratic has exactly two roots in total, so the only possibilities are two real roots, one repeated real root or none.
Which method should I use first in an exam?: Try factorisation first, because it is fastest when b2 - 4ac is a perfect square like 81. If no factor pair appears within about a minute, switch to the formula, which always works.