Solving Quadratic Equations | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

What you are solving

A can be written as $ax^2+bx+c=0$ , where $a\ne0$ .
A makes the equation true. On its graph, real roots are the $x$ -intercepts. “Real” means the root is an ordinary number on the number line.
An expression has no equals sign. An equation has an equals sign. A function gives one output, often written $y=f(x)$ , for each allowed input $x$ .

The roots of x² + 5x + 6 = 0 are the two places where the curve meets the line y = 0.

Expression → rewrite itEquation → solve for xFunction → read its graph

The calculator-first method

Unless a method is named, put the equation in zero form (with zero alone on one side) and use the polynomial solver first.

Write $ax^2+bx+c=0$ . A missing term has coefficient zero. A coefficient is the signed number multiplying a term.
fx-991EX: MENU → Equation/Func → Polynomial → degree 2. On an fx-991CW: HOME → Equation → Polynomial → ax² + bx + c = 0.
Enter signed $a$ , $b$ , $c$ . Record both roots.

Worked example

Solve $6x^2+5=17x$

Rearrange and enter $a=6$ , $b=-17$ , $c=5$ .
$6x^2-17x+5=0$
The solver gives the convenient roots $x=\tfrac52$ and $x=\tfrac13$ .
Turn the roots into factors and show the working.
$(2x-5)(3x-1)=0$

Answer

x=\tfrac52 \quad\text{or}\quad x=\tfrac13

A root $x=\tfrac pq$ gives the factor $qx-p$ . This is because $qx-p=0$ gives $x=\tfrac pq$ . Expand once to check the coefficient of $x^2$ .

Use the roots to choose the working

Nice roots → write factorsAwkward roots → use the formulaRepeated root → state it onceNo real roots → state this clearlyNamed method → follow it

Worked example

Solve $6x^2-3x-2=0$ , giving the roots to 3 significant figures

The solver gives $0.879\ldots$ and $-0.379\ldots$ . They are awkward, so use the formula.
Use $a=6$ , $b=-3$ , $c=-2$ .
$\begin{aligned}x&=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\dfrac{3\pm\sqrt{57}}{12}\end{aligned}$

Answer

\begin{aligned}x&=0.879\\\text{or}\quad x&=-0.379\end{aligned}

For an exact answer, keep the square root in $x=\dfrac{3\pm\sqrt{57}}{12}$ instead of changing it to a decimal. For 3 significant figures, count from the first non-zero digit and round only the final answer.

Keep every valid root

Show the zero form and one supporting factorisation or formula line.
Keep both roots, then apply any original restriction or context.
Check by expanding the factors or substituting into the original equation.

Common mistake

Dividing

x^2=5x

x

loses

x=0

. Write

x(x-5)=0

, so

x=0

x=5

Your turnMethod selection practice

Use the solver first, then show suitable working.

$2x^2+7x+3=0$
$5x^2+3x-1=0$ , giving exact roots

Show worked answer

1. Convenient roots: write factors.

(2x+1)(x+3)=0

x=-\tfrac12 \quad\text{or}\quad x=-3

2. Awkward roots: use the formula.

x=\dfrac{-3\pm\sqrt{3^2-4(5)(-1)}}{2(5)}

Answer

\begin{aligned}x&=\dfrac{-3+\sqrt{29}}{10}\\\text{or}\quad x&=\dfrac{-3-\sqrt{29}}{10}\end{aligned}

What you are solving

A can be written as $ax^2+bx+c=0$ , where $a\ne0$ .
A makes the equation true. On its graph, real roots are the $x$ -intercepts. “Real” means the root is an ordinary number on the number line.
An expression has no equals sign. An equation has an equals sign. A function gives one output, often written $y=f(x)$ , for each allowed input $x$ .

The roots of x² + 5x + 6 = 0 are the two places where the curve meets the line y = 0.

Expression → rewrite itEquation → solve for xFunction → read its graph

The calculator-first method

Unless a method is named, put the equation in zero form (with zero alone on one side) and use the polynomial solver first.

Write $ax^2+bx+c=0$ . A missing term has coefficient zero. A coefficient is the signed number multiplying a term.
fx-991EX: MENU → Equation/Func → Polynomial → degree 2. On an fx-991CW: HOME → Equation → Polynomial → ax² + bx + c = 0.
Enter signed $a$ , $b$ , $c$ . Record both roots.

Worked example

Solve $6x^2+5=17x$

Rearrange and enter $a=6$ , $b=-17$ , $c=5$ .
$6x^2-17x+5=0$
The solver gives the convenient roots $x=\tfrac52$ and $x=\tfrac13$ .
Turn the roots into factors and show the working.
$(2x-5)(3x-1)=0$

Answer

x=\tfrac52 \quad\text{or}\quad x=\tfrac13

A root $x=\tfrac pq$ gives the factor $qx-p$ . This is because $qx-p=0$ gives $x=\tfrac pq$ . Expand once to check the coefficient of $x^2$ .

Use the roots to choose the working

Nice roots → write factorsAwkward roots → use the formulaRepeated root → state it onceNo real roots → state this clearlyNamed method → follow it

Worked example

Solve $6x^2-3x-2=0$ , giving the roots to 3 significant figures

The solver gives $0.879\ldots$ and $-0.379\ldots$ . They are awkward, so use the formula.
Use $a=6$ , $b=-3$ , $c=-2$ .
$\begin{aligned}x&=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\dfrac{3\pm\sqrt{57}}{12}\end{aligned}$

Answer

\begin{aligned}x&=0.879\\\text{or}\quad x&=-0.379\end{aligned}

Keep every valid root

Show the zero form and one supporting factorisation or formula line.
Keep both roots, then apply any original restriction or context.
Check by expanding the factors or substituting into the original equation.

Common mistake

Dividing

x^2=5x

x

loses

x=0

. Write

x(x-5)=0

, so

x=0

x=5

Your turnMethod selection practice

Use the solver first, then show suitable working.

$2x^2+7x+3=0$
$5x^2+3x-1=0$ , giving exact roots

Show worked answer

1. Convenient roots: write factors.

(2x+1)(x+3)=0

x=-\tfrac12 \quad\text{or}\quad x=-3

2. Awkward roots: use the formula.

x=\dfrac{-3\pm\sqrt{3^2-4(5)(-1)}}{2(5)}

Answer

\begin{aligned}x&=\dfrac{-3+\sqrt{29}}{10}\\\text{or}\quad x&=\dfrac{-3-\sqrt{29}}{10}\end{aligned}