6.02d, 6.03b - Solving general quadratics | OCR GCSE Mathematics

What can you do when a quadratic has no convenient integer factors? Under 6.03b, you will rearrange and solve general quadratics by factorisation, completing the square or the quadratic formula; 6.02d adds recognising and using that formula. You will keep exact roots when needed and understand why a quadratic can have two, one or no real roots.

Using the quadratic formula

Can you solve $2x^2+3x-4=0$ without guessing its factors?

Higher only. Factorising is efficient when suitable factors are easy to find. The quadratic formula gives another route for every real quadratic $ax^2+bx+c=0$ , with $a\ne0$ , as long as you interpret the square root correctly. First rearrange to zero, then identify the signed coefficients. For the opening equation they are $a=2$ , $b=3$ , $c=-4$ .

Quadratic formula

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

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

For real roots, the expression under the square root must be non-negative. Use the plus and minus branches separately; when that expression is zero, they give the same root. The quadratic formula is supplied on the June/November 2027 Higher formula sheet. Learn its structure and how to use it.

The denominator $2a$ divides the whole numerator. Use brackets around a negative coefficient when substituting, especially in $b^2$ and $-4ac$ .

Worked example 1: a negative constant

Question. Solve $2x^2+3x-4=0$ , giving exact answers and approximations to three decimal places.

Plan. Substitute $a=2$ , $b=3$ , $c=-4$ , simplify the discriminant, then calculate each branch.

\begin{aligned} x&=\frac{-3\pm\sqrt{3^2-4(2)(-4)}}{2(2)} &&\text{substitute signed coefficients}\\ &=\frac{-3\pm\sqrt{41}}4 &&\text{simplify inside the root}\\ x&\approx0.851\quad\text{or}\quad x\approx-2.351 &&\text{round only at the end} \end{aligned}

Answer. $x=(-3+\sqrt{41})/4$ or $x=(-3-\sqrt{41})/4$ , approximately $0.851$ and $-2.351$ .

Check. For either exact root, $4x+3=\pm\sqrt{41}$ , so $16x^2+24x+9=41$ ; dividing the resulting $16x^2+24x-32=0$ by eight recovers the original equation.

Change only the sign of the linear coefficient. Its square stays nine, but the $-b$ term changes sign.

Worked example 2: a negative linear coefficient

Question. Solve $2x^2-3x-4=0$ , giving exact answers and approximations to three decimal places.

Plan. Use $b=-3$ with brackets, and keep both branches.

\begin{aligned} x&=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(-4)}}{2(2)} &&\text{signed substitution}\\ &=\frac{3\pm\sqrt{41}}4 &&\text{simplify}\\ x&\approx2.351\quad\text{or}\quad x\approx-0.851 &&\text{evaluate both branches} \end{aligned}

Answer. $x=(3\pm\sqrt{41})/4$ , approximately $2.351$ or $-0.851$ .

Check. $4x-3=\pm\sqrt{41}$ gives $16x^2-24x-32=0$ , eight times the original. The roots are also the negatives of those in example 1: replacing $x$ by $-x$ changes only the linear term's sign.

That sign connection is useful: changing $b$ to $-b$ in $ax^2+bx+c=0$ negates every root. It provides a second check independent of a calculator entry.

Watch out

For $b=-3$ , $-b=3$ and $b^2=9$ . Writing $b^2=-9$ changes the formula. Also, $(-b+\sqrt{D})/(2a)$ requires the denominator under the whole numerator.

Keep both exact branches in your answer.

Solving by completing the square

Completing the square rewrites an equation so that one squared bracket can be isolated. It is especially useful when the linear coefficient is easy to halve.

You already know $x^2+4x=(x+2)^2-4$ . In an equation, once the square is isolated, a positive right side gives two square-root branches. Keep the outside coefficient when completing a non-monic square.

Worked example 3: an outside coefficient

Question. Solve $2x^2+8x+3=0$ by completing the square. Give exact answers.

Plan. Factor two from the first two terms, complete the inner square, then take both roots.

\begin{aligned} 2\bigl((x+2)^2-4\bigr)+3&=0 &&\text{complete the square}\\ 2(x+2)^2-5&=0 &&\text{combine constants}\\ (x+2)^2&=\frac52 &&\text{add 5, divide by 2}\\ x&=-2\pm\frac{\sqrt{10}}2 &&\text{take both square roots} \end{aligned}

Answer. $x=-2+\sqrt{10}/2$ or $x=-2-\sqrt{10}/2$ .

Check. Either root gives $(x+2)^2=10/4=5/2$ . Thus $2(x+2)^2-5=0$ ; expanding that left side gives $2x^2+8x+3$ .

Change only the original constant from three to five. The midpoint of the two roots stays at $-2$ , but their distance from it changes.

Worked example 4: the same square, a new constant

Question. Solve $2x^2+8x+5=0$ by completing the square. Give exact answers.

Plan. Use the same squared bracket and calculate its new required value.

\begin{aligned} 2\bigl((x+2)^2-4\bigr)+5&=0 &&\text{complete the same square}\\ 2(x+2)^2-3&=0 &&\text{combine constants}\\ (x+2)^2&=\frac32 &&\text{isolate the square}\\ x&=-2\pm\frac{\sqrt6}{2} &&\text{take both roots} \end{aligned}

Answer. $x=-2\pm\sqrt6/2$ .

Check. Squaring $x+2=\pm\sqrt6/2$ gives $3/2$ ; then $2(x+2)^2-3=0$ , which expands to the original equation. Subtracting two from a positive square root does not remove the negative branch.

This method also explains the formula. For $ax^2+bx+c=0$ , $a\ne0$ , multiply by $4a$ and add $b^2$ :

\begin{aligned} 4a^2x^2+4abx&=-4ac\\ (2ax+b)^2&=b^2-4ac\\ 2ax+b&=\pm\sqrt{b^2-4ac}\qquad(b^2-4ac\ge0). \end{aligned}

Subtract $b$ and divide by the nonzero $2a$ to obtain the quadratic formula. The formula is completed-square reasoning packaged for repeated use.

Watch out

From $(x+2)^2=3/2$ , write $x+2=\pm\sqrt{3/2}$ , then subtract two. Writing $x=\pm\sqrt{3/2-2}$ puts the subtraction inside the root and changes the equation.

Use the new constant to find the new distance from the same midpoint.

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6.02d, 6.03b - Solving general quadratics

Using the quadratic formula

Solving by completing the square

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