FA15, HA18 - Solving quadratics by factorising

FA15, HA18 - Solving quadratics by factorising

Learn why a product equal to zero gives separate equations, how to find every solution of a factorisable quadratic, and how to check the roots. This develops FA15 and HA18 using monic quadratics already written equal to zero; the factorisation makes their solutions visible.

Zero products

How can (x−2)(x+5)(x-2)(x+5) equal zero when the two brackets cannot both be zero at the same value of xx? Only one factor needs to be zero.

For numbers, 0×7=00\times7=0 and (−4)×0=0(-4)\times0=0. But multiplying two non-zero real numbers cannot give zero: if ab=0ab=0 and a≠0a\ne0, dividing by aa gives b=0b=0.

Zero-product rule

If a product of real numbers is zero, at least one factor is zero. Thus ab=0ab=0 gives a=0a=0 or b=0b=0, possibly both. It does not require both to be zero.

Apply the rule to each bracket, then solve the resulting linear equations. “Or” means either value can make the original product zero.

Worked example 1: one zero bracket at a time

Question. Solve (x−2)(x+5)=0(x-2)(x+5)=0.

Plan. Set each factor equal to zero separately.

x−2=0orx+5=0zero-product rulex=2orx=−5solve each equation\begin{aligned} x-2=0&\quad\text{or}\quad x+5=0 &&\text{zero-product rule}\\ x=2&\quad\text{or}\quad x=-5 &&\text{solve each equation} \end{aligned}

Answer. x=2x=2 or x=−5x=-5.

Check. At x=2x=2, the product is 0×7=00\times7=0. At x=−5x=-5, it is (−7)×0=0(-7)\times0=0. Both work.

Change only +5+5 to −5-5. The second solution changes sign because it must make its bracket zero.

Worked example 2: reading the signs correctly

Question. Solve (x−2)(x−5)=0(x-2)(x-5)=0.

Plan. Write both zero-factor equations rather than copy the bracket numbers.

x−2=0orx−5=0at least one factor is zerox=2orx=5add 2 or 5\begin{aligned} x-2=0&\quad\text{or}\quad x-5=0 &&\text{at least one factor is zero}\\ x=2&\quad\text{or}\quad x=5 &&\text{add }2\text{ or }5 \end{aligned}

Answer. x=2x=2 or x=5x=5.

Check. The products are 0×(−3)=00\times(-3)=0 at x=2x=2 and 3×0=03\times0=0 at x=5x=5.

The zero on the right matters. From ab=6ab=6, you cannot conclude a=6a=6 or b=6b=6: 2×3=62\times3=6 is a counterexample. This lesson uses quadratic equations already equal to zero, so the zero-product rule applies directly after factorising.

Watch out

Do not divide (x−2)(x+5)=0(x-2)(x+5)=0 by x−2x-2: that division is invalid at x=2x=2, and it loses a solution. Split into zero-factor equations instead.

Find both values for a new pair of brackets.

Finding both solutions

For a quadratic given in expanded form, factorising is the step that creates a product to which the zero-product rule applies.

The expression x2+bx+cx^2+bx+c is a sum, so it is not valid to set each of its three terms equal to zero. First find two bracket numbers with sum bb and product cc. Then solve the two linear equations.

Worked example 3: factorise, then solve

Question. Solve x2+3x−18=0x^2+3x-18=0.

Plan. Factorise using a pair with product −18-18 and sum 33, then set each bracket to zero.

x2+3x−18=(x+6)(x−3)6×(−3)=−18, 6−3=3(x+6)(x−3)=0same equation in product formx+6=0orx−3=0zero-product rulex=−6orx=3solve\begin{aligned} x^2+3x-18&=(x+6)(x-3) &&6\times(-3)=-18,\ 6-3=3\\ (x+6)(x-3)&=0 &&\text{same equation in product form}\\ x+6=0&\quad\text{or}\quad x-3=0 &&\text{zero-product rule}\\ x=-6&\quad\text{or}\quad x=3 &&\text{solve} \end{aligned}

Answer. x=−6x=-6 or x=3x=3.

Check. In the original equation, 36−18−18=036-18-18=0 at x=−6x=-6, and 9+9−18=09+9-18=0 at x=3x=3.

Now change only the sign of the middle coefficient. The required product stays the same; the signed pair changes.

Worked example 4: a negative middle coefficient

Question. Solve x2−3x−18=0x^2-3x-18=0.

Plan. Choose numbers with product −18-18 and sum −3-3.

x2−3x−18=(x−6)(x+3)(−6)×3=−18, −6+3=−3(x−6)(x+3)=0use the factorisationx−6=0orx+3=0set factors to zerox=6orx=−3solve\begin{aligned} x^2-3x-18&=(x-6)(x+3) &&(-6)\times3=-18,\ -6+3=-3\\ (x-6)(x+3)&=0 &&\text{use the factorisation}\\ x-6=0&\quad\text{or}\quad x+3=0 &&\text{set factors to zero}\\ x=6&\quad\text{or}\quad x=-3 &&\text{solve} \end{aligned}

Answer. x=6x=6 or x=−3x=-3.

Check. 36−18−18=036-18-18=0 at x=6x=6; 9+9−18=09+9-18=0 at x=−3x=-3. Each value satisfies the original equation.

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