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This page sets out the method for the quadratic formula in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the The quadratic formula topic page.
When the quadratic is already , or after you have rearranged it.
When you see , , or anything not already .
If ax² + bx + c = 0 will not factor with integers, use x = (-b ± √(b² - 4ac)) / (2a).
b² - 4ac tells you the roots: positive gives two real roots, zero gives one, negative gives none.
If b² - 4ac is not a square number, the roots stay in surd form. Do not replace √ with a decimal.
(2 ± 2√2) / 2 is 1 ± √2. Divide the whole numerator and the denominator by the same number.
The denominator is 2a. When a is 2 or 3, do not secretly divide by 2 as if a were 1.
The ± sign is two calculations. Write both, unless the discriminant is zero.
Start from zero on one side. x^2+4x=1 must become x^2+4x-1=0 before you read c=-1. Leaving the 1 on the right and then using c=1 changes the discriminant from 20 to 12 and both answers are wrong. For 2x^2+5x-3=0, write a=2, b=5, c=-3 on one line. Brackets around a negative c stop the sign of -4ac from flipping in your head.
The formula has -b, not b. If b=-5, then -b=5. Students who type the coefficient as it stands, minus sign and all, into the "-b" place end up subtracting when they should add. Say it as "the opposite of b". For x^2-5x+6=0, b=-5, so -b=5, and the discriminant is (-5)^2-4(1)(6)=25-24=1. The solutions are (5± 1)/2, which are 3 and 2.
Method: 1. Arrange the equation as ax^2+bx+c=0. 2. Read a, b and c with their signs. 3. Write -b and b^2 before you touch the formula. 4. Keep the fraction bar over the whole numerator.
Calculate b^2-4ac on its own line. For 2x^2+5x-3=0, that is 25-4(2)(-3)=25+24=49. A perfect square means the quadratic factorises, and the formula should agree with the factors. (-5± 7)/4 gives 2/4=1/2 and (-12)/4=-3. The factors (2x-1)(x+3) give the same solutions. If your formula answers do not match a factorisation you trust, the discriminant sign is the first place to look.
A negative discriminant is an answer, not a failed calculation. For x^2+x+1=0, D=1-4=-3. Stop and write "no real solutions". Do not invent a square root of a negative number at this level. A zero discriminant means the two branches of the ± meet, so there is one repeated solution. Show D=0 and then the single value of x. Leaving ± 0 in the answer looks unfinished.
Method: 1. Work out b^2-4ac on its own line. 2. A perfect square gives rational solutions. 3. Zero gives one repeated solution. 4. A negative value means no real solution. Stop and say so.
For x^2+4x-1=0, a=1, b=4, c=-1 and D=16+4=20. Then x=dfrac-4±√(20)2. Simplify the root before you divide: √(20)=√(4× 5)=2√(5), so x=dfrac-4± 2√(5)2=-2±√(5). Dividing only the root by 2a, and leaving -4 untouched, is the error that produces -4±√(5). The bar is under both parts. Cancel the 2 in every term.
You can check a surd solution without a calculator by rearranging. If x=-2+√(5), then x+2=√(5) and (x+2)^2=5. Expanding gives x^2+4x+4=5, so x^2+4x-1=0. The same check works for -2-√(5) because the square removes the sign. A decimal check is fine on a calculator paper, but keep at least four decimal places until the final rounding. Rounding √(20) to 4.5 before dividing moves the answer off the mark.
Method: 1. Simplify the square root before you divide, when it has a square factor. 2. Divide the whole numerator by 2a, not only one term. 3. Leave an exact answer in surd form when the question asks for exact. 4. Check that both the plus and the minus have been written if there are two roots.
The ± means you write two solutions unless the discriminant is zero. x=1±√(5) is an acceptable exact pair. If the question says "to 2 decimal places", calculate each branch separately and round at the end. Do not round one branch and then use that rounded number to make the other. And do not give only the positive-looking branch: both are required unless the question asks for a positive length or a similar restriction.
For 2x^2+3x-7=0, D=9+56=65, which does not simplify, so x=dfrac-3±√(65)4. State a, b and c, show the discriminant, and keep the fraction. A coordinate or length question may then ask you to reject one solution. Reject it in a sentence after you have found both, for example "length is positive, so take the positive root". Crossing out the other root with no reason loses the interpretation mark.
Method: 1. The ± is two solutions unless the discriminant is zero. 2. Write both, or write the ± form if that was requested. 3. Substitute one root back if you have time. 4. Do not round an exact answer unless the question says to.
The formula only works on an equation written as ax^2+bx+c=0, with zero on the right. Exam questions often hide the standard form. In x(2x+5)=9, expand the bracket to get 2x^2+5x=9, and then subtract 9 from both sides to get 2x^2+5x-9=0. Now a=2, b=5 and c=-9. Forgetting to subtract and using c=9 is the usual way to lose all the marks, because every later line is built on the wrong number.
Some equations need more work. For 3x^2+x=5, subtract 5 to get 3x^2+x-5=0. For (x+1)(x+2)=7, expand to x^2+3x+2=7 and subtract 7 to get x^2+3x-5=0. In each case write the standard form on its own line and then list a, b and c with their signs. If a question says "give your answers to 2 decimal places" or "3 significant figures", it is a clue that the quadratic does not factorise.
Method: 1. Expand any brackets and collect all terms on one side so the other side is zero. 2. Write the equation as ax^2+bx+c=0. 3. List a, b and c with their signs, including any minus signs. 4. Look at the wording of the question for the accuracy needed.
Tip: If the question asks for decimal answers, the quadratic is not meant to factorise, so go straight to the formula.
The formula is x=dfrac-b±√(b^2-4ac)2a. When you type it, use brackets around the whole numerator and around the whole denominator. For 2x^2+5x-9=0 type (-5+√(5^2-4× 2×(-9)))÷(2× 2). Without the brackets on the bottom the calculator divides by 2 and then multiplies by 2, which is a different number. Put brackets around negative numbers that are squared, so that (-9)^2 gives 81 and not -81.
Work out the part under the root first and write it down: 25+72=97. Then do the two calculations, one with + and one with -, so that you get both solutions. For 2x^2+5x-9=0 they are x=1.2122ldots and x=-3.7122ldots, which are 1.21 and -3.71 to 2 decimal places. Always check one answer by putting it back into the original equation, using the unrounded value, and confirming it gives a number very close to zero.
Method: 1. Work out b^2-4ac first and write the value down. 2. Type the numerator in brackets and divide by the denominator in brackets. 3. Calculate once with plus and once with minus. 4. Round only at the end and check by substituting back.
When a question says "leave your answer in surd form", do not use decimals. For x^2-8x+9=0 the discriminant is 64-36=28. Simplify √(28) as √(4× 7)=2√(7). Then x=dfrac8±2√(7)2. Every term on top has a factor of 2 that matches the denominator, so divide each term: x=4±√(7). Divide every term of the numerator, not just one.
Check numerically: 4+√(7)=6.6458ldots and substituting into the equation gives 44.166-53.166+9=0. Another form asks for answers like -5±2√(3) for x^2+10x+13=0. Here b^2-4ac=100-52=48=16× 3, so √(48)=4√(3), and dfrac-10±4√(3)2=-5±2√(3). Look for the largest square factor in the discriminant, which saves a step.
Method: 1. Work out the discriminant and find its largest square factor. 2. Write the root as a whole number times a smaller surd. 3. Divide every term in the numerator by any common factor with the denominator. 4. Give the answer as two separate solutions, or as a ± expression.
The expression under the root, b^2-4ac, tells you how many real solutions there are without solving. If it is positive there are two different solutions. If it is zero there is exactly one repeated solution. If it is negative there are no real solutions, because you cannot take the square root of a negative number. To show that x^2+3x+5=0 has no real solutions, work out 3^2-4× 1× 5=-11 and state that it is negative.
Questions at the top of the paper put an unknown in the equation. If x^2+6x+k=0 has equal roots, then 36-4k=0, so k=9. If x^2-4x+k=0 has two different real solutions, then 16-4k>0, so k<4. Test your answer with a number: k=3 gives 16-12=4, which is positive, as expected, and k=5 gives -4, which is negative. Write the discriminant as a single expression first, then solve the equation or inequality.
Method: 1. Write a, b and c, including the unknown, and form b^2-4ac. 2. Set it equal to zero for equal roots, greater than zero for two roots, or less than zero for none. 3. Solve the resulting equation or inequality for the unknown. 4. Test a value either side to confirm the direction of the inequality.
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