E2.5 - Solving quadratic equations (Extended)

E2.5 - Solving quadratic equations (Extended)

Two numbers can satisfy the same quadratic equation. How can you find both, and what can you do when no integer factor pair works? This Extended E2.5 lesson develops factorisation and the supplied quadratic formula, including exact answers written with surds.

Factorisation

Which numbers make x2−7x+6=0x^2-7x+6=0? Trying a few integers finds one answer, but factorisation exposes every possibility: x2−7x+6=(x−1)(x−6)x^2-7x+6=(x-1)(x-6).

Root of an equation

A root, or solution, is a value of the unknown that makes the equation true. A quadratic equation can be written as ax2+bx+c=0ax^2+bx+c=0, where a≠0a\ne0.

Two non-zero real numbers cannot multiply to give zero. Therefore, once a product equals zero, at least one of its factors must be zero. This turns one quadratic equation into two linear possibilities.

Zero-product property

AB=0⟺A=0 or B=0.AB=0\quad\Longleftrightarrow\quad A=0\text{ or }B=0.

The product must equal zero before you use this property.

Not on the supplied formula lists: learn it.

First put every term on one side, factorise fully, and set each variable factor equal to zero. For example, x2−7x=−6x^2-7x=-6 becomes x2−7x+6=0x^2-7x+6=0 by adding 6 to both sides. The word “or” matters: the two factors do not have to be zero at the same time.

Worked example 1: two roots from two factors

Question. Solve x2−7x+6=0x^2-7x+6=0.

Plan. Factorise using two numbers with product 6 and sum −7-7, then use the zero-product property.

x2−7x+6=(x−1)(x−6)constants −1 and −6(x−1)(x−6)=0x−1=0orx−6=0\begin{aligned} x^2-7x+6&=(x-1)(x-6) &&\text{constants }-1\text{ and }-6\\ (x-1)(x-6)&=0\\ x-1=0&\quad\text{or}\quad x-6=0 \end{aligned}

Answer. x=1x=1 or x=6x=6.

Check. At x=1x=1, 1−7+6=01-7+6=0. At x=6x=6, 36−42+6=036-42+6=0. Both satisfy the original equation.

Change only the coefficient of x2x^2. A factor now starts with 2x2x, so one root is a fraction; setting a factor to zero still works in exactly the same way.

Worked example 2: a non-monic quadratic

Question. Solve 2x2−7x+6=02x^2-7x+6=0.

Plan. Split the middle term and solve the two linear factors; the equation already has zero on one side.

2x2−4x−3x+6=0split −7x2x(x−2)−3(x−2)=0(2x−3)(x−2)=02x−3=0orx−2=0\begin{aligned} 2x^2-4x-3x+6&=0 &&\text{split }-7x\\ 2x(x-2)-3(x-2)&=0\\ (2x-3)(x-2)&=0\\ 2x-3=0&\quad\text{or}\quad x-2=0 \end{aligned}

Answer. x=32x=\frac32 or x=2x=2.

Check. In the original left side, x=32x=\frac32 gives 92−212+6=0\frac92-\frac{21}2+6=0; x=2x=2 gives 8−14+6=08-14+6=0. Both equal the original right side.

A zero constant needs care: x2−7x=x(x−7)x^2-7x=x(x-7), so x2−7x=0x^2-7x=0 has roots 0 and 7. Dividing by xx would discard the possible root 0. A repeated factor such as (x−3)2=0(x-3)^2=0 gives just one distinct root, x=3x=3.

Watch out

You cannot set each factor to zero in (x−2)(x+3)=6(x-2)(x+3)=6. Subtract 6 and rearrange first. Factorisation solves a zero product, not an arbitrary product.

Keep both possibilities in your answer.

Quadratic formula

Factorisation is convenient when the factors are visible. The quadratic formula works from the three coefficients, so you do not need to guess factors. It is a hand method as well as a useful calculator-paper method.

Quadratic formula

For ax2+bx+c=0ax^2+bx+c=0, where a≠0a\ne0,

x=−b±b2−4ac2a.x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

The symbol ±\pm means calculate once with plus and once with minus. Over the real numbers, a positive value inside the square root gives two distinct roots, zero gives a repeated root, and a negative value gives no real roots.

Supplied on page 2 of Extended Papers 2 and 4; all required formulas are given within questions on Papers 5 and 6.

Write the equation in the required form before identifying aa, bb and cc. Include each coefficient's sign. The complete numerator, including both −b-b and the square root, is divided by 2a2a.

Worked example 3: a negative constant

Question. Use the quadratic formula to solve 3x2+5x−2=03x^2+5x-2=0.

Plan. Substitute a=3a=3, b=5b=5, c=−2c=-2, keeping the negative constant in brackets.

x=−5±52−4(3)(−2)2(3)=−5±25+246subtracting a negative adds=−5±76\begin{aligned} x&=\frac{-5\pm\sqrt{5^2-4(3)(-2)}}{2(3)}\\ &=\frac{-5\pm\sqrt{25+24}}6 &&\text{subtracting a negative adds}\\ &=\frac{-5\pm7}6 \end{aligned}

Answer. x=13x=\frac13 or x=−2x=-2.

Check. 3(13)2+5(13)−2=13+53−2=03(\frac13)^2+5(\frac13)-2=\frac13+\frac53-2=0. Also, 3(−2)2+5(−2)−2=12−10−2=03(-2)^2+5(-2)-2=12-10-2=0.

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