2.7 Quadratic equations

Learning objectives

2.7A Factorising simple quadratic equationssolve quadratic equations by factorisation, limited to x² + bx + c = 0 2.7HA Factorising quadratic equationssolve quadratic equations by factorisation 2.7HB Quadratic formula and completing the squaresolve quadratic equations by using the quadratic formula or completing the square 2.7HC Forming quadratic equationsform and solve quadratic equations from data given in a context 2.7HD Linear and quadratic simultaneous equationssolve simultaneous equations in two unknowns, one linear and one quadratic

To solve a monic quadratic by factorisation, first write it in the form $x^2+bx+c=0$ , factorise the left side, then use the zero-product rule.

Step	Action
standardise	move all terms to one side so the other side is 0
choose	find $p,q$ with $pq=c$ and $p+q=b$
factor	write $(x+p)(x+q)=0$
solve	set each factor equal to 0
check	substitute both roots into the original equation

For $x^2-5x-36=0$ , use $4$ and $-9$ : $(x+4)(x-9)=0$ . Hence $x=-4$ or $x=9$ .

The zero-product rule works because a product is zero only when at least one factor is zero. It applies after the equation has been set equal to zero.

Do not stop at the factorised expression or report only one root. A quadratic can have two distinct roots, one repeated root, or no real factor pair.

A general quadratic $ax^2+bx+c=0$ may have a leading coefficient other than 1. Factor out any common factor, then construct two linear factors whose product recreates all three terms.

Structure	Factorisation route
common factor	remove it first
$A^2-B^2$	use $(A-B)(A+B)$
$ax^2+bx+c$	find terms with product $ac$ and sum $b$ , split the middle term, group
already a product equal to a value	expand/rearrange to make the equation equal 0

Solve $6x^2-x-2=0$ . Since $ac=-12$ and $3+(-4)=-1$ , write $6x^2+3x-4x-2=0$ , so $(3x-2)(2x+1)=0$ . Thus $x=2/3$ or $x=-1/2$ .

Expand the factors before applying the zero-product rule. The leading, middle and constant coefficients must exactly match the standardised equation.

Factorisation solves only when the equation is zero. From $(x+1)(3x-2)=5$ , neither factor can be set to 5 or 0 until the equation is rearranged correctly.

For $ax^2+bx+c=0$ with $a
e0$, the quadratic formula solves every quadratic; completing the square exposes the same roots through a squared expression.

Method	Core form	Best use
quadratic formula	$x=(-b\pm\sqrt{b^2-4ac})/(2a)$	reliable for any coefficients
completing the square	$a(x-h)^2+k=0$	reveals symmetry and roots together
discriminant	$D=b^2-4ac$	predicts the number of real roots

For $2x^2+3x-1=0$ , $a=2,b=3,c=-1$ , so $x=[-3\pm\sqrt{17}]/4$ . Keep the entire numerator over $2a$ .

For $x^2-6x+5=0$ , write $(x-3)^2-4=0$ . Then $(x-3)^2=4$ , so $x=3\pm2$ , giving $1$ and $5$ .

If $D>0$ there are two real roots; if $D=0$ one repeated real root; if $D<0$ no real roots. An exact surd should not be rounded unless requested.

The $\pm$ belongs before the whole square root and the denominator is $2a$ . Omitting either sign loses a root.

A contextual quadratic comes from expressing related lengths, areas, products or other quantities in one unknown, imposing the stated condition, then solving and interpreting the roots.

Stage	Control question
define	what does the unknown represent and what units apply?
express	how is every related quantity written using it?
connect	which area, product, total or equality creates the equation?
solve	can it be factorised, or is another quadratic method needed?
interpret	which roots satisfy lengths, counts and original restrictions?

A trapezium has parallel sides $x+5$ and $3x-2$ , height $2x-3$ , and area 133. Then $[(x+5)+(3x-2)](2x-3)/2=133$ , which simplifies to $8x^2-6x-275=0$ .

Solve the derived equation, then return to every original expression. Reject a root that makes a length non-positive, violates a denominator restriction, or conflicts with the stated domain.

A valid algebraic root is not automatically a valid contextual answer. State the requested quantity, not merely the value of an auxiliary variable.

A linear and a quadratic equation can meet at up to two points. Use the linear equation to express one variable, substitute into the quadratic, then recover and correctly pair both coordinates.

Step	Action
isolate	make $x$ or $y$ the subject of the linear equation
substitute	replace that variable everywhere in the quadratic equation
solve	simplify to one quadratic and find all roots
recover	substitute each root into the linear relation
pair	report each matching $(x,y)$ solution and verify both equations

For $y=3-2x$ and $x^2+y^2=18$ , substitution gives $x^2+(3-2x)^2=18$ , hence $5x^2-12x-9=0=(5x+3)(x-3)$ . The solutions are $(-0.6,4.2)$ and $(3,-3)$ .

The solutions are intersection points of a line and a conic. Two roots mean two intersections, a repeated root means tangency, and no real roots means no real intersection.

Do not mix the recovered values. Each root must be substituted separately and paired with its own corresponding value.

Next Topic

—2.8 InequalitiesRead

2.7 Quadratic equations

Learning objectives

Solve monic quadratics by factorisation

Solve general quadratics by factorisation

Use the quadratic formula or complete the square

Form and solve a quadratic from context

Solve linear–quadratic simultaneous equations

Next Topic