P1 1.5 - Solving quadratic equations

P1 1.5 - Solving quadratic equations

Solve quadratic equations by factorisation, completing the square, the quadratic formula and a calculator. Choose a method that suits the equation, retain every solution and distinguish an exact answer from a rounded approximation.

Factorisation and zero products

If x2=4xx^2=4x, dividing both sides by xx gives x=4x=4. But x=0x=0 also satisfies the original equation. How can we solve it without losing that answer?

Move every term to one side, so the other side is zero. Then factorise rather than dividing by an expression that might itself be zero.

Zero-product rule

For real numbers u,vu,v,

uv=0⟺u=0 or v=0.uv=0\quad\Longleftrightarrow\quad u=0\text{ or }v=0.

A nonzero factor cannot turn another nonzero factor into zero. The rule applies to a product equal to zero, not a product equal to a different number.

Not in the formula booklet: learn it.

Solving a quadratic means finding every value that makes its equation true. Those values are its roots; geometrically, they are the xx-coordinates where its zero-form graph meets the xx-axis.

Worked example 1: retaining a zero root

Solve x2−4x=0x^2-4x=0.

Plan. Factor out xx, then set each factor equal to zero.

x(x−4)=0factor the left sidex=0orx−4=0use the zero-product rule\begin{aligned} x(x-4)&=0 &&\text{factor the left side}\\ x=0\quad&\text{or}\quad x-4=0 &&\text{use the zero-product rule} \end{aligned}

Answer. x=0x=0 or x=4x=4.

Check. 02−4(0)=00^2-4(0)=0 and 42−4(4)=16−16=04^2-4(4)=16-16=0. These are the two solutions of the opening equation.

Change only the right side from 00 to 55. Now rearrangement must come before the zero-product rule.

Worked example 2: make a zero side first

Solve x2−4x=5x^2-4x=5.

Plan. Subtract 55, factorise the resulting quadratic, then solve both linear factors.

x2−4x−5=0subtract 5 from both sides(x−5)(x+1)=0product −5, sum −4x−5=0orx+1=0.\begin{aligned} x^2-4x-5&=0 &&\text{subtract }5\text{ from both sides}\\ (x-5)(x+1)&=0 &&\text{product }-5,\text{ sum }-4\\ x-5=0\quad&\text{or}\quad x+1=0. \end{aligned}

Answer. x=5x=5 or x=−1x=-1.

Check. In the original equation, 25−20=525-20=5 and 1+4=51+4=5.

The first step decides whether the product rule is available.

Watch out

From x(x−4)=5x(x-4)=5, you cannot set x=5x=5 or x−4=5x-4=5. A product can equal 55 in many ways. Rearrange to a product equal to zero instead.

Retain both roots, including any fractional one.

Completing the square to solve

A quadratic that does not factorise neatly can still become a squared bracket equal to a number.

You already know how to complete the square. For solving, rearrange until the squared bracket is alone, then take both square roots. If u2=d>0u^2=d>0, both u=du=\sqrt d and u=−du=-\sqrt d work. If d=0d=0, they coincide; if d<0d<0, there is no real solution.

Worked example 3: an irrational pair of roots

Solve x2−6x+2=0x^2-6x+2=0, giving exact answers.

Plan. Complete the square, isolate it and use both signs of the square root.

(x−3)2−7=0complete the square(x−3)2=7add 7x−3=±7take both rootsx=3±7add 3\begin{aligned} (x-3)^2-7&=0 &&\text{complete the square}\\ (x-3)^2&=7 &&\text{add }7\\ x-3&=\pm\sqrt7 &&\text{take both roots}\\ x&=3\pm\sqrt7 &&\text{add }3 \end{aligned}

Answer. x=3+7x=3+\sqrt7 or x=3−7x=3-\sqrt7.

Check. Substitute x=3+tx=3+t into the original expression: it becomes t2−7t^2-7. For either t=7t=\sqrt7 or −7-\sqrt7, this is zero.

Change only the right side from 00 to 1010. The completed square stays the same, but the number it equals changes.

Worked example 4: a nonzero right side

Solve x2−6x+2=10x^2-6x+2=10, giving exact answers.

Plan. Use the same completed form, then add 77 to both sides.

(x−3)2−7=10(x−3)2=17x−3=±17x=3±17.\begin{aligned} (x-3)^2-7&=10\\ (x-3)^2&=17\\ x-3&=\pm\sqrt{17}\\ x&=3\pm\sqrt{17}. \end{aligned}

Answer. x=3+17x=3+\sqrt{17} or x=3−17x=3-\sqrt{17}.

Check. Substituting x=3+tx=3+t gives t2−7=17−7=10t^2-7=17-7=10 for both signs. The roots are equally spaced about x=3x=3, the graph's symmetry line.

The square-root symbol itself is nonnegative; the two solutions appear because of the separate plus and minus signs.

Watch out

u2=9u^2=9 gives u=±3u=\pm3. Writing 9=±3\sqrt9=\pm3 confuses the two statements: 9\sqrt9 alone means 33.

Complete a square before taking a root.

Quadratic formula and exact roots

The quadratic formula packages the completed-square method into one result that works for any quadratic equation.

Starting from ax2+bx+c=0ax^2+bx+c=0, where a≠0a\ne0, completing the square gives

(x+b2a)2=b2−4ac4a2.\left(x+\frac b{2a}\right)^2=\frac{b^2-4ac}{4a^2}.

Taking both signs and rearranging yields the formula below. When a<0a<0, the denominator's sign swaps which root is called the plus root, but the same pair is obtained.

Quadratic formula

For ax2+bx+c=0ax^2+bx+c=0, with real coefficients and a≠0a\ne0,

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

The roots are real when the quantity under the square root is nonnegative.

Not in the formula booklet: learn it.

Write the coefficients with their signs before substituting. The entire numerator is divided by 2a2a.

Worked example 5: a positive constant

Solve 2x2−4x+1=02x^2-4x+1=0, giving exact answers.

Plan. Use a=2a=2, b=−4b=-4, c=1c=1, then simplify the surd and fraction.

x=−(−4)±(−4)2−4(2)(1)2(2)=4±84evaluate the square-root argument=4±224=1±22simplify both numerator terms\begin{aligned} x&=\frac{-(-4)\pm\sqrt{(-4)^2-4(2)(1)}}{2(2)}\\ &=\frac{4\pm\sqrt8}{4} &&\text{evaluate the square-root argument}\\ &=\frac{4\pm2\sqrt2}{4}=1\pm\frac{\sqrt2}{2} &&\text{simplify both numerator terms} \end{aligned}

Answer. x=1±2/2x=1\pm\sqrt2/2.

Check. The original polynomial is 2(x−1)2−12(x-1)^2-1. At either proposed root, (x−1)2=1/2(x-1)^2=1/2, so it is zero.

Change only the sign of the constant term. Its sign must remain attached to cc in the formula.

Worked example 6: a negative constant

Solve 2x2−4x−1=02x^2-4x-1=0, giving exact answers.

Plan. Keep a=2a=2, b=−4b=-4, but now use c=−1c=-1.

x=4±16−4(2)(−1)4=4±244=4±264=1±62.\begin{aligned} x&=\frac{4\pm\sqrt{16-4(2)(-1)}}4\\ &=\frac{4\pm\sqrt{24}}4\\ &=\frac{4\pm2\sqrt6}{4}=1\pm\frac{\sqrt6}{2}. \end{aligned}

Subtracting a negative product makes the square-root argument 2424, not 88.

Answer. x=1±6/2x=1\pm\sqrt6/2.

Check. The polynomial is 2(x−1)2−32(x-1)^2-3. Each root makes (x−1)2=3/2(x-1)^2=3/2, giving zero.

An exact answer should remain exact until a decimal is requested.

Watch out

The denominator 2a2a divides both −b-b and the square-root term. For example, (4+24)/4(4+\sqrt{24})/4 is not 4+24/44+\sqrt{24}/4.

Use signed coefficients and simplify the root.

Calculator roots and checking

A calculator can supply roots efficiently, but you still need to enter the correct equation, record both answers and choose the requested precision.

Arrange the equation as ax2+bx+c=0ax^2+bx+c=0, then use the quadratic-equation facility with coefficients a,b,ca,b,c. Alternatively, evaluate the plus and minus versions of the quadratic formula separately, using brackets around the whole numerator and denominator. Menu names vary, but the mathematical input is the same. Unless the question states otherwise, give inexact P1 answers to 33 significant figures and retain working precision before that final rounding.

Worked example 7: calculator coefficients in zero form

Solve 1.2x2−3x−2=01.2x^2-3x-2=0, giving your answers to 33 significant figures.

Plan. Enter a=1.2a=1.2, b=−3b=-3, c=−2c=-2, then retain both displayed roots before rounding.

x=3±(−3)2−4(1.2)(−2)2(1.2)=3±18.62.4.x=\frac{3\pm\sqrt{(-3)^2-4(1.2)(-2)}}{2(1.2)} =\frac{3\pm\sqrt{18.6}}{2.4}.

The unrounded values are approximately 3.0469882213.046988221 and −0.546988221-0.546988221.

Answer. x=3.05x=3.05 or x=−0.547x=-0.547, to 33 significant figures.

Check. Substituting the rounded values into the left side gives 0.0130.013 and 0.00005080.0000508, close to zero; using unrounded roots reduces these residuals to calculator rounding error. At x=3x=3, the polynomial is negative and at x=3.1x=3.1 it is positive, consistent with the positive root.

Change only the right side from 00 to 44. The calculator's constant coefficient must now change.

Worked example 8: rearrangement before entry

Solve 1.2x2−3x−2=41.2x^2-3x-2=4, giving your answers to 33 significant figures.

Plan. Subtract 44 before entering the equation, so the constant is −6-6.

1.2x2−3x−6=0x=3±9+28.82.4=3±37.82.4.\begin{aligned} 1.2x^2-3x-6&=0\\ x&=\frac{3\pm\sqrt{9+28.8}}{2.4} =\frac{3\pm\sqrt{37.8}}{2.4}. \end{aligned}

The roots are approximately 3.8117376923.811737692 and −1.311737692-1.311737692.

Answer. x=3.81x=3.81 or x=−1.31x=-1.31, to 33 significant figures.

Check. In the original left side, the rounded values give 3.989323.98932 and 3.989323.98932, close to the required 44. Checking against zero before rearranging would test the wrong statement.

The opening problem shows why checking both roots matters: a method can produce a correct answer while missing another.

Exam tip

If the question asks for exact roots, a rounded decimal is not a replacement for a surd. If it requests a stated accuracy, round the final roots rather than the square-root argument or an intermediate fraction.

Use a calculator, but show which polynomial you solved.

Quick Recap

  • Factorise a zero-form equation and set each factor equal to zero.
  • Completing the square isolates a square; remember both square-root signs.
  • The quadratic formula works for any quadratic with nonzero leading coefficient.
  • Calculator entry needs signed coefficients in zero form, both roots and correct final rounding.

For the opening equation, factorising x(x−4)=0x(x-4)=0 keeps both x=0x=0 and x=4x=4.