A ladder leans against the back wall of a bakery's delivery bay, resting on level ground, as shown. The foot of the ladder is \(x\) m from the wall, the lad...
Question 1 Report
A ladder leans against the back wall of a bakery's delivery bay, resting on level ground, as shown. The foot of the ladder is \(x\) m from the wall, the ladder is \((x+7)\) m long, and it reaches a point \((x+1)\) m up the wall.
- Show that \(x^2 - 12x - 48 = 0\). (2)
- Solve this equation to find \(x\), giving your answer to \(3\) significant figures. (2)
Answer Details
This question models a ladder against a wall as a right-angled triangle with algebraic side lengths, uses Pythagoras' theorem to form a quadratic equation, and then solves it, rejecting the negative root.
The ladder is the hypotenuse, length \((x+7)\) m, the foot of the ladder is \(x\) m from the wall, and the ladder reaches \((x+1)\) m up the wall. By Pythagoras' theorem:
\[(x+7)^2 = x^2+(x+1)^2\]Expanding both sides:
\[x^2+14x+49 = x^2+x^2+2x+1\]Subtracting \(x^2\) and rearranging:
\[x^2-12x-48=0\]as required [2 marks].
Using the quadratic formula with \(a=1, b=-12, c=-48\):
\[x = \dfrac{12 \pm \sqrt{144+192}}{2} = \dfrac{12 \pm \sqrt{336}}{2} = 6 \pm 2\sqrt{21}\]Since \(x\) is a distance, the negative root \(6-2\sqrt{21}\) is rejected, giving:
\[x = 6+2\sqrt{21} = 15.2 \text{ (3 s.f.)}\][2 marks]
Expanding \((x+7)^2\) and \((x+1)^2\) carefully term by term avoids the common error of dropping the middle terms \(14x\) and \(2x\), which would otherwise leave a linear rather than a quadratic equation.
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