A18 - Quadratic formula and completing the square (Higher)

A18 - Quadratic formula and completing the square (Higher)

Some quadratics do not split into brackets with whole-number coefficients. You will solve them using the quadratic formula or completing the square, keep exact surd answers, and give rounded answers when asked. You will also put equations into a suitable form before choosing a method.

Using the quadratic formula (Higher)

The equation x2−6x+1=0x^2-6x+1=0 has roots between 00 and 11, and between 55 and 66. How can you find them exactly when whole-number factor pairs fail? Keep that puzzle: a square and a square root will unlock it.

You already know

A quadratic has the form ax2+bx+c=0ax^2+bx+c=0, with a≠0a\ne0. Factorising can give exact roots. Surds keep irrational square roots exact, and completing the square rewrites a quadratic without changing its value.

Start with x2+2x−1=0x^2+2x-1=0. No pair of integers has product −1-1 and sum 22, so the usual integer factor search stalls. The quadratic formula works whether or not such a pair exists. The signs belong to the coefficients: here a=1a=1, b=2b=2 and c=−1c=-1.

Quadratic formula

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

x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Use the coefficients after the equation has been written with zero on one side. The sign ±\pm means work out the plus and minus possibilities.

On the 2027 Higher formulae sheet.

The whole numerator is divided by 2a2a. If b2−4ac>0b^2-4ac>0, there are two distinct real roots. If it equals zero, the two answers coincide. If it is negative, there are no real roots, because no real number has a negative square. These root-count facts are not on the formulae sheet: learn them.

Worked example 1: an exact surd answer

Question. Solve x2+2x−1=0x^2+2x-1=0. Give your answers in exact form. Non-calculator.

Plan. Substitute the signed coefficients into the formula, then simplify the square root.

a=1,b=2,c=−1read the coefficientsx=−2±22−4(1)(−1)2(1)substitute=−2±82simplify inside the root=−1±28=22\begin{aligned} a&=1,\quad b=2,\quad c=-1&&\text{read the coefficients}\\ x&=\frac{-2\pm\sqrt{2^2-4(1)(-1)}}{2(1)}&&\text{substitute}\\ &=\frac{-2\pm\sqrt8}{2}&&\text{simplify inside the root}\\ &=-1\pm\sqrt2&&\sqrt8=2\sqrt2 \end{aligned}

Why keep c=−1c=-1 in brackets? Subtracting 4ac4ac here adds 44; losing that negative changes the roots.

Answer. x=−1+2x=-1+\sqrt2 or x=−1−2x=-1-\sqrt2.

Check. Put x=−1+sx=-1+s, where s=±2s=\pm\sqrt2. Then x2+2x−1=s2−2=0x^2+2x-1=s^2-2=0.

Change only the constant term from −1-1 to −2-2. This time the question asks for decimal answers, so use a calculator after writing the exact expression.

Worked example 2: rounding both roots

Question. Solve x2+2x−2=0x^2+2x-2=0. Give your answers to 33 decimal places. Calculator allowed.

Plan. Use the formula, simplify exactly, then evaluate the two signs separately.

x=−2±22−4(1)(−2)2(1)substitute=−2±122=−1±3simplifyx1=−1+3=0.732050807…evaluate plusx2=−1−3=−2.732050807…evaluate minus\begin{aligned} x&=\frac{-2\pm\sqrt{2^2-4(1)(-2)}}{2(1)}&&\text{substitute}\\ &=\frac{-2\pm\sqrt{12}}2=-1\pm\sqrt3&&\text{simplify}\\ x_1&=-1+\sqrt3=0.732050807\ldots&&\text{evaluate plus}\\ x_2&=-1-\sqrt3=-2.732050807\ldots&&\text{evaluate minus} \end{aligned}

Why round at the end? An early approximation to 3\sqrt3 can change the final requested decimal place.

Answer. x=0.732x=0.732 or x=−2.732x=-2.732, to 33 decimal places.

Check. The exact roots give x2+2x−2=0x^2+2x-2=0. Using 0.7320.732 gives −0.000176-0.000176, a small rounding difference.

For a negative bb, square the whole signed number: (−5)2=25(-5)^2=25. Enter the numerator and denominator in separate calculator brackets, for example (−b+b2−4ac)/(2a)(-b+\sqrt{b^2-4ac})/(2a), then repeat with the minus sign.

Watch out

Wrong: x=−b+b2−4ac/2ax=-b+\sqrt{b^2-4ac}/2a, with only the root divided. Right: x=−b+b2−4ac2ax=\dfrac{-b+\sqrt{b^2-4ac}}{2a}. The fraction bar groups the entire numerator and the entire denominator.

Keep the middle coefficient and change the constant once more.

Solving by completing the square (Higher)

Completing the square replaces a quadratic by a single squared bracket and a constant. Once the square is alone, taking both square roots gives the solutions.

For the opening equation, x2−6x+1=(x−3)2−8x^2-6x+1=(x-3)^2-8. This is the rewriting you used to locate turning points. Now set the expression equal to zero and solve it. The identity x2+bx=(x+b/2)2−(b/2)2x^2+bx=(x+b/2)^2-(b/2)^2 and this solving method are not on the formulae sheet: learn them.

Worked example 3: solving a completed square

Question. Solve x2−6x+1=0x^2-6x+1=0 by completing the square. Give exact answers. Non-calculator.

Plan. Rewrite using (x−3)2(x-3)^2, isolate the square, then take both roots.

(x−3)2−9+1=0complete the square(x−3)2=8add 8x−3=±8take both square rootsx=3±22simplify and add 3\begin{aligned} (x-3)^2-9+1&=0&&\text{complete the square}\\ (x-3)^2&=8&&\text{add }8\\ x-3&=\pm\sqrt8&&\text{take both square roots}\\ x&=3\pm2\sqrt2&&\text{simplify and add }3 \end{aligned}

Why both square roots? Either 222\sqrt2 or −22-2\sqrt2 has square 88.

Answer. x=3+22x=3+2\sqrt2 or x=3−22x=3-2\sqrt2.

Check. Each answer makes (x−3)2−8=8−8=0(x-3)^2-8=8-8=0, so it solves the original equation too.

Change only the scale: double every term in the starting equation. Divide out that non-zero common factor before completing the square.

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