Solving Quadratics by Factorising (Ex 5G)

Year 10 Mathematics Core · Cambridge Ch 5, §5G · Mr Wong

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Learning intentions.

Use the Null Factor Law: if $p\times q=0$, then $p=0$ or $q=0$
Rearrange to $ax^2+bx+c=0$, factorise, then solve
Factorise with a common factor, a difference of two squares, and monic / non-monic trinomials

Standard form is $ax^2+bx+c=0$. Difference of two squares: $a^2-b^2=(a-b)(a+b)$. A quadratic can have $0$, $1$ or $2$ solutions.

Warm-up — already factorised

1. State the solutions.

a $x(x-6)=0$

b $(x-5)(x+1)=0$

c $(2x-1)(x+4)=0$

d $(3x+2)(4x-3)=0$

2. Solve each equation.

a $x^2+5x=0$

b $x^2-49=0$

c $3x^2=75$

d $9x^2-25=0$

3. Solve by factorising.

a $x^2+7x+10=0$

b $x^2-9x+20=0$

c $x^2+2x-15=0$

d $x^2+6x+9=0$

Part C (non-monic) · Part D (disguised quadratics) · Challenge

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4. Solve by splitting the middle term.

a $2x^2+5x-3=0$

b $3x^2-10x+8=0$

5. Rearrange into standard form first, then solve.

a $x^2=x+30$

b $x^2=4(x+15)$

6. Solve $\dfrac{x+12}{x}=x$. Show every step, and state how many solutions the equation has.