When should you reach for the quadratic formula rather than factorising?

Whenever the answers are wanted to a stated accuracy, such as two decimal places, since that tells you they are not whole numbers or simple fractions. It is also faster on awkward coefficients, and unlike factorising it always works .

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Quadratic Functions (Edexcel IGCSE Further Pure Maths): Flashcards

Exam code: 4PM1

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FrontFactorising Quadratics
How do you factorise a quadratic with no constant term, such as $5 x^{2} + 30 x$ ?

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How do you factorise a quadratic with no constant term, such as $5 x^{2} + 30 x$ ?
Take out the highest common factor, which will always include $x$ itself.
Here that factor is $5 x$ , giving $5 x (x + 6)$ .
To factorise $x^{2} + b x + c$ as $(x + p) (x + q)$ , complete what the two numbers must do:
$p q =______and p + q =______$
The completed conditions are:
$p q = c and p + q = b$
So for $x^{2} - 21 x - 100$ the numbers are $4$ and $- 25$ , giving $(x + 4) (x - 25)$ .
How do you factorise $12 x^{2} - 11 x - 5$ by grouping?
Find two numbers multiplying to $a c = - 60$ and adding to $b = - 11$ , which are $4$ and $- 15$ , then use them to split the middle term.
Grouping the pairs gives $4 x (3 x + 1) - 5 (3 x + 1)$ , and the common bracket factors out as $(4 x - 5) (3 x + 1)$ .
When is grouping better than factorising a quadratic by inspection?
When the coefficient of $x^{2}$ is not $1$ , so there is no single pair of brackets to spot directly.
Inspection can still work on those, but grouping is reliable: it follows the same rule every time instead of depending on trial and error.
Define difference of two squares.
A difference of two squares is any expression of the form $a^{2} - b^{2}$ , one square subtracted from another.
It always factorises as $(a + b) (a - b)$ , and $a$ and $b$ need not be single letters: $9 x^{2} - 16$ is ${(3 x)}^{2} - 4^{2}$ , so it factorises as $(3 x + 4) (3 x - 4)$ .
True or False?
$x^{2} + 36$ can be factorised as a difference of two squares.
False.
A difference of two squares needs a subtraction, and $x^{2} + 36$ is a sum, so it does not factorise at all.
Compare $x^{2} - 36$ , which does fit the pattern and factorises as $(x + 6) (x - 6)$ .
Define completing the square.
Completing the square means rewriting a quadratic in a form containing a squared bracket, usually $a {(x + p)}^{2} + q$ .
The variable then appears once instead of twice, which is what makes the rest of the information about the curve readable straight off.
Complete the rule for the first two terms of a quadratic, where $p$ is half of $b$ :
$x^{2} + b x = {(x +______)}^{2} -______$
The completed rule is:
$x^{2} + b x = {(x + p)}^{2} - p^{2}$
Squaring the bracket produces an extra $p^{2}$ that was not there before, so subtracting it puts the expression back to what it was.
What happens to the rule when $b$ is negative, as in $x^{2} - 20 x$ ?
Nothing changes: $p$ is still half of $b$ , so $p = - 10$ and $x^{2} - 20 x = {(x - 10)}^{2} - 100$ .
The term at the end stays negative whatever the sign of $b$ , because $p^{2}$ is a square.
How do you complete the square on $4 x^{2} + 16 x + 5$ ?
Factor the $4$ out of the $x^{2}$ and $x$ terms only, leaving the constant outside: $4 [x^{2} + 4 x] + 5$ .
Complete the square inside, then multiply back out: $4 [{(x + 2)}^{2} - 4] + 5 = 4 {(x + 2)}^{2} - 11$ .
Where is the turning point of $y = a {(x + p)}^{2} + q$ ?
At $(- p, q)$ , and note the negative sign in the $x$ coordinate.
It is a minimum point when $a > 0$ and a maximum point when $a < 0$ .
True or False?
Every quadratic graph has exactly one turning point.
True.
Completing the square rewrites any quadratic as $a {(x + p)}^{2} + q$ , and that bracket can only be zero at one value of $x$ .
Everywhere else the squared term contributes something positive, so that single point is the one extreme value of the curve.
How does completing the square show that $x^{2} + 6 x + 11$ is always positive?
Completed, the expression is ${(x + 3)}^{2} + 2$ , and a squared bracket can never be negative.
The smallest value ${(x + 3)}^{2}$ can take is $0$ , so the whole expression is never less than $2$ .
Complete the quadratic formula for the solutions of $a x^{2} + b x + c = 0$ :
$x = \frac{- b \pm \sqrt{______-______}}{______}$
The completed formula is:
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
The $2 a$ divides the whole numerator, the $- b$ included, which is why the $- b$ sits inside the fraction rather than in front of it.
When should you reach for the quadratic formula rather than factorising?
Whenever the answers are wanted to a stated accuracy, such as two decimal places, since that tells you they are not whole numbers or simple fractions.
It is also faster on awkward coefficients, and unlike factorising it always works.
How do you solve $x^{2} - 4 x = 0$ ?
Take out the common factor to get $x (x - 4) = 0$ , giving $x = 0$ or $x = 4$ .
Never divide through by $x$ to simplify it first: that would throw away the solution $x = 0$ altogether.
True or False?
If a quadratic expression will not factorise, the equation has no solutions.
False.
Failing to factorise usually just means the solutions are not whole numbers or simple fractions: $x^{2} - 7 x + 2 = 0$ does not factorise, but has solutions $6.70$ and $0.30$ to two decimal places.
A quadratic with no real solutions does also fail to factorise, so the two cases overlap, but they are not the same thing.
After completing the square, what step turns ${(x + 3)}^{2} = 4$ into two solutions?
Take the square root of both sides and write $\pm$ in front of the root: $x + 3 = \pm 2$ .
Forgetting the $\pm$ loses a solution; here $x = - 3 + 2 = - 1$ or $x = - 3 - 2 = - 5$ .
How do you make $x$ the subject of $x^{2} + 6 x = y$ ?
Complete the square first, so that $x^{2} + 6 x = {(x + 3)}^{2} - 9$ and the equation becomes ${(x + 3)}^{2} - 9 = y$ .
Then rearrange and take roots, keeping the $\pm$ : ${(x + 3)}^{2} = y + 9$ , so $x = - 3 \pm \sqrt{y + 9}$ .
Define the discriminant of a quadratic.
For $a x^{2} + b x + c$ with $a \neq 0$ , the discriminant is $Δ = b^{2} - 4 a c$ .
It is exactly the expression that sits under the square root in the quadratic formula.
What do $Δ > 0$ , $Δ = 0$ and $Δ < 0$ tell you about the roots?
There are three cases:
- $Δ > 0$ : unequal real roots, two distinct solutions
- $Δ = 0$ : equal real roots, one repeated solution
- $Δ < 0$ : the roots are not real
If a question says only that the roots are real, without saying how many, use $Δ \geq 0$ .
What does the discriminant tell you about the graph of $y = a x^{2} + b x + c$ ?
How often the curve meets the $x$ -axis: twice if $Δ > 0$ , once if $Δ = 0$ and never if $Δ < 0$ .
When $Δ = 0$ the $x$ -axis is a tangent to the curve, and when $Δ < 0$ the whole curve lies either above or below the axis.
How do you find the values of $k$ for which $2 k x^{2} + k x - k + 2 = 0$ has two distinct roots?
Write down the discriminant with $a = 2 k$ , $b = k$ and $c = - k + 2$ , then set it greater than zero.
That gives $9 k^{2} - 16 k > 0$ , and solving that inequality gives $k < 0$ or $k > \frac{16}{9}$ .
For $a x^{2} + b x + c = 0$ with roots $α$ and $β$ , complete the two results:
$α + β =______and α β =______$
The completed results are:
$α + β = - \frac{b}{a} and α β = \frac{c}{a}$
Note the minus sign on the sum but not on the product, which is the commonest slip here.
How do you write a quadratic equation whose roots have sum $S$ and product $P$ ?
Use $x^{2} - S x + P = 0$ , which comes from expanding $(x - α) (x - β)$ .
Multiply through by whatever clears the fractions when integer coefficients are asked for, taking the multiplier from the denominators of $S$ and $P$ .
How are $α^{2} + β^{2}$ and ${(α - β)}^{2}$ written in terms of $α + β$ and $α β$ ?
Both come from squaring the sum:
$α^{2} + β^{2} = {(α + β)}^{2} - 2 α β and {(α - β)}^{2} = {(α + β)}^{2} - 4 α β$
Squaring $α + β$ introduces an extra $2 α β$ , so subtract $2 α β$ for the sum of squares and $4 α β$ to turn that middle term negative.
True or False?
Two different quadratic equations can have exactly the same pair of roots.
True.
Multiplying an equation through by any non-zero number leaves its roots untouched, so $2 x^{2} - 7 x - 4 = 0$ and $4 x^{2} - 14 x - 8 = 0$ have the same two roots.
That is why a question asking you to form an equation specifies integer coefficients, and you generally give the simplest such equation.

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