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Quadratic Equations formulas

11 formulas and 24 common traps for SPI Mathematics Quadratic Equations, grouped by subtopic.

Full notes with worked examples

Recognising a Quadratic Equation

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Standard form of a quadratic equation

Standard form

ax2+bx+c=0,a≠0ax^2 + bx + c = 0, \quad a \ne 0

Writing a word situation as a quadratic equation

Pythagoras theorem

(hypotenuse)2=(base)2+(height)2(\text{hypotenuse})^2 = (\text{base})^2 + (\text{height})^2

Common traps

Thinking b or c must be non-zero

Only a must be non-zero. x2−16=0x^2 - 16 = 0 has b=0b = 0 and is still a quadratic equation of degree 2.

Keeping the sign when a term crosses the = sign

A term changes sign when it moves to the other side: x2=3x+4x^2 = 3x + 4 becomes x2−3x−4=0x^2 - 3x - 4 = 0, not x2+3x+4=0x^2 + 3x + 4 = 0.

Judging by the look before expanding

An x2x^2 on both sides proves nothing. x(x+3)=x2−5x(x + 3) = x^2 - 5 loses its x2x^2 terms and reduces to 3x+5=03x + 5 = 0, a linear equation.

Confusing a root in the coefficient with a root of x

3x2+x−1=0\sqrt{3}x^2 + x - 1 = 0 is quadratic, because 3\sqrt{3} is just a number. x2+x=1x^2 + \sqrt{x} = 1 is not, because x itself is under the root.

Adding the hypotenuse to another side

The hypotenuse stands alone on one side: hypotenuse2=a2+b2\text{hypotenuse}^2 = a^2 + b^2. An equation that adds the hypotenuse squared to another side squared is wrong.

Reading 'more than' backwards

'6 more than twice x' is 2x+62x + 6, and '2 less than h' is h−2h - 2. Write the base quantity first, then add or take away.

Solving by Factorisation and the Formula

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A root of a quadratic equation makes it true

Condition for a root

aα2+bα+c=0a\alpha^2 + b\alpha + c = 0
  • ααa root of ax² + bx + c = 0

Solving a quadratic equation by splitting the middle term

Zero product rule

(x−p)(x−q)=0⇒x=p or x=q(x - p)(x - q) = 0 \Rightarrow x = p \text{ or } x = q

The quadratic formula

Quadratic formula

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Common traps

Squaring a negative number wrongly

(−2)2=4(-2)^2 = 4, not −4-4. Put a negative root in brackets before squaring it.

Dividing both sides by x

Dividing x2=7xx^2 = 7x by x leaves only x=7x = 7 and loses the root 0. Move everything to one side and factorise: x(x−7)=0x(x - 7) = 0.

Taking the root with the bracket's sign

The factor (x+3)(x + 3) gives the root −3-3, and (x−3)(x - 3) gives +3+3. The root has the opposite sign to the number in the bracket.

Factorising before one side is 0

In x2+3x=10x^2 + 3x = 10, move the 10 first: x2+3x−10=0x^2 + 3x - 10 = 0, which is (x+5)(x−2)=0(x + 5)(x - 2) = 0.

Losing the sign of b

The formula starts with −b-b. When b=−4b = -4, −b=+4-b = +4. Write b with its sign before you start.

Dividing only part of the top by 2a

4±252=2±5\frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}: both the 4 and the 252\sqrt{5} are divided by 2, not just one of them.

The Discriminant and Nature of Roots

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Nature of the roots from the discriminant

Discriminant

D=b2−4acD = b^2 - 4ac

Values of k that give real, distinct or no real roots

Conditions on the discriminant

b2−4ac≥0 (real roots)b2−4ac>0 (distinct roots)b2−4ac<0 (no real roots)b^2 - 4ac \ge 0 \text{ (real roots)} \qquad b^2 - 4ac > 0 \text{ (distinct roots)} \qquad b^2 - 4ac < 0 \text{ (no real roots)}

Equal roots: setting the discriminant to zero

Equal roots

b2−4ac=0x=−b2ab^2 - 4ac = 0 \qquad x = -\frac{b}{2a}

Common traps

Dropping the sign of c

When c is negative, −4ac-4ac is positive. For 3x2+x−2=03x^2 + x - 2 = 0, D=1+24=25D = 1 + 24 = 25, not 1−241 - 24.

Calling equal roots rational

D = 0 means the roots are equal, not that they are rational. x2−22x+2=0x^2 - 2\sqrt{2}x + 2 = 0 has D = 0 and both roots equal to 2\sqrt{2}, which is irrational.

Keeping only one side of a square inequality

b2<25b^2 < 25 does not mean just b<5b < 5: b=−7b = -7 gives 49. The answer is −5<b<5-5 < b < 5.

Using > when the question says real

Real roots include equal roots, so the condition is D≥0D \ge 0. Only 'distinct' or 'unequal' roots need D>0D > 0.

Taking only the positive square root

k2=25k^2 = 25 gives k=5k = 5 or k=−5k = -5. An option with only one sign is incomplete when both values work.

Reading b before expanding

In x2+k(2x−3)=0x^2 + k(2x - 3) = 0 the coefficient of x is 2k2k and the constant is −3k-3k. Expand first, then read a, b and c.

Ignoring 'positive' in 'equal and positive roots'

D = 0 often gives two values of the unknown. Find the equal root −b2a-\frac{b}{2a} for each and keep only the value that makes it positive.

Sum and Product of the Roots

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Sum and product of the roots from the coefficients

Sum and product of roots

α+β=−baαβ=ca\alpha + \beta = -\frac{b}{a} \qquad \alpha\beta = \frac{c}{a}

Forming a quadratic equation from its roots

Equation from its roots

x2−(α+β)x+αβ=0x^2 - (\alpha + \beta)x + \alpha\beta = 0

Rebuilding an equation from wrongly copied coefficients

Correct equation from the two students' trusted values

x2−(sum from the student with the right x term) x+(product from the student with the right constant)=0x^2 - (\text{sum from the student with the right } x \text{ term})\,x + (\text{product from the student with the right constant}) = 0

Common traps

Forgetting the minus in the sum

The sum of the roots is −ba-\frac{b}{a}. For x2−9x+18=0x^2 - 9x + 18 = 0 it is +9+9, and for x2+9x+18=0x^2 + 9x + 18 = 0 it is −9-9.

Not dividing by a

For 2x2−8x+3=02x^2 - 8x + 3 = 0 the product of the roots is 32\frac{3}{2}, not 3. Divide by a unless a is 1.

Writing plus the sum

The equation is x2−(sum)x+product=0x^2 - (\text{sum})x + \text{product} = 0. Roots 2 and 5 give x2−7x+10=0x^2 - 7x + 10 = 0; x2+7x+10=0x^2 + 7x + 10 = 0 has the roots −2-2 and −5-5.

Leaving fractions in the equation

Options are printed with whole-number coefficients. x2−52x+1=0x^2 - \frac{5}{2}x + 1 = 0 appears as 2x2−5x+2=02x^2 - 5x + 2 = 0, so multiply through before comparing.

Trusting the wrong student's sum

A student who copied the coefficient of x wrongly has a wrong sum of roots. Take only the product from that student, and the sum from the one who got the x term right.

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