Quadratic Equations and Inequalities Worked ExamplesIGCSE Maths 0580

Find two numbers multiplying to $+12$ and adding to $-7$. Both must be negative.

Write the factorised form.

A product is zero only if one of its factors is zero, so set each bracket to zero.

Solve each.

Rearrange so that one side is zero. Move the terms to the side with the $x^2$ term so that its coefficient stays positive.

Multiply first and last coefficients, then find two numbers with that product summing to $-5$.

Split the middle term.

Factorise in pairs.

Take out the common bracket.

Set each factor to zero.

Identify the coefficients, keeping the signs.

Substitute into the formula.

Evaluate the discriminant. Note $-4 \times 2 \times -7 = +56$.

Substitute back.

Evaluate both roots and round as instructed.

Halve the coefficient of $x$ and square it: $\left(\tfrac{6}{2}\right)^2 = 9$.

Substitute into the equation.

Simplify and isolate the square.

Take the square root of both sides — both signs are needed.

Subtract $3$.

Area is length times width.

Expand the left side.

Rearrange so the equation equals zero.

Factorise — two numbers multiplying to $-56$ and adding to $1$.

$x = -8$ would give a negative width, so it is rejected. Use $x = 7$.

Factorise the quadratic to find where it equals zero.

These are the critical values. Sketch the curve: the $x^2$ coefficient is positive, so it is a U-shape crossing at $-2$ and $3$.

'$> 0$' means where the curve is above the $x$-axis — the two outer regions, not the middle.

Test a value in each region: $x = 0$ gives $-6$ (not $> 0$), $x = 4$ gives $6$ ✓, $x = -3$ gives $6$ ✓.

(a) The longest side, $(x + 8)$, is opposite the right angle, so it is the hypotenuse. Apply Pythagoras' theorem.

Expand both brackets fully — each squared bracket gives three terms.

Collect every term on the left. The $x^{2}$ terms leave $x^{2}$, the $x$ terms leave $-2x$ and the constants leave $-15$. This is the given equation, so (a) is shown.

(b) Factorise: two numbers multiplying to $-15$ and adding to $-2$ are $-5$ and $+3$.

$x$ is a length, so $x = -3$ is rejected. With $x = 5$ the sides are $5$, $12$ and $13\text{ cm}$ (check: $25 + 144 = 169$ ✓).

(a) '$< 0$' means the part of the curve below the $x$-axis. The curve crosses the axis at $x = -1$ and $x = 3$, and is below it between these values. The inequality is strict, so the end values are not included.

(b) Draw the line $y = 5$. It meets the curve at $x = -2$ and $x = 4$. '$\geqslant 5$' means the curve is on or above this line, which happens on the two outer branches.

Check algebraically: subtract $5$ from both sides and factorise. The critical values match the graph.

(c) The drawn curve is $y = x^{2} - 2x - 3$, so rewrite the new inequality to match it by subtracting $3$ from both sides.

Draw the line $y = -3$. It meets the curve at $x = 0$ and $x = 2$, and the curve is below it between them.

(a) The border appears on both sides of the picture. The outside width is $30 + x + x = 30 + 2x$, and the outside height is $20 + 2x + 2x = 20 + 4x$. Sam has only added one border to each dimension. (Sam's right-hand side, $1200$, is correct: picture plus an equal frame is $2 \times 600$.)

(b) The whole outside rectangle has area picture $+$ frame $= 600 + 600 = 1200$.

Expand the brackets.

Rearrange to $= 0$, then divide every term by $8$ to reach the given equation.

(c) The quadratic does not factorise, so use the formula with $a = 1$, $b = 20$, $c = -75$.

This gives $x = 3.2287\ldots$ or $x = -23.228\ldots$. A width cannot be negative, so reject the second value.

(d) Use the unrounded value of $x$ for the dimensions, then round to $3$ significant figures.

(a) Time $=$ distance $\div$ speed, so the outward crossing takes $\frac{30}{v}$ hours and the return takes $\frac{30}{v - 5}$ hours. The return is longer by $20$ minutes, which is $\frac{1}{3}$ of an hour — convert before writing the equation.

Multiply every term by $3v(v - 5)$ to clear all three denominators — including the $\frac{1}{3}$ on the right.

Expand both sides. The $90v$ terms cancel on the left.

Collect everything on the side with the $v^{2}$ term. This is the given equation.

(b) Use the formula with $a = 1$, $b = -5$, $c = -450$. Because $b$ is negative, $-b = +5$, and $(-5)^{2} = 25$.

This gives $v = 23.860\ldots$ or $v = -18.860\ldots$. A speed cannot be negative, so reject the second value. Keep the full value for part (c).

(c) The return speed is $v - 5 = 18.860\ldots\text{ km/h}$.

The whole-number part is $1$ hour. Convert the decimal part to minutes by multiplying by $60$.

(a) When $x = 5$, the bracket $x + 2$ equals $7$. Squaring gives the same result for $+7$ and $-7$, so the other solution makes the bracket equal to $-7$. No formula is needed.

(Check with symmetry: the two roots are equally spaced either side of $x = -2$. $5$ is $7$ above $-2$, and $-9$ is $7$ below it.)

(b) Substitute the known solution into the equation.

The quadratic does not factorise, and the answer is asked for in surd form, so use the formula with $a = 2$, $b = 6$, $c = -3$. The numerator starts with $-b = -6$.

Evaluate the discriminant carefully: $-4 \times 2 \times (-3) = +24$, so it is $36 + 24$, not $36 - 24$.

Simplify the surd using the largest square factor of $60$.

Every term in the numerator and denominator is divisible by $2$, so cancel.