A rectangular garden bed measures \((2x+1)\) m by \((x-3)\) m and has an area of 60 m\(^2\). (a) Show that \(2x^2-5x-63=0\). [2] (b) Solve this equation to ...

An area given as a product of two linear expressions leads to a quadratic. Part (a) asks for the quadratic to be established, and because the result is printed, every algebraic step must be shown. Part (b) then solves it and converts the answer back into a length.

(a) Area is length \(\times\) width, so

\((2x + 1)(x - 3) = 60\) [M1]

Expanding the left side: \(2x^2 - 6x + x - 3 = 2x^2 - 5x - 3\), so the equation is \(2x^2 - 5x - 3 = 60\). Subtracting \(60\) from both sides gives

\(2x^2 - 5x - 63 = 0\) [A1]

(b) Two numbers multiplying to \(2 \times (-63) = -126\) and adding to \(-5\) are \(-14\) and \(9\), giving \(2x^2 - 14x + 9x - 63 = 2x(x - 7) + 9(x - 7)\), so

\((2x + 9)(x - 7) = 0\), so \(x = 7\), and the length is \(2(7) + 1 = 15\) m [B1]

The other root, \(x = -4.5\), is rejected because it would make the width \(x - 3\) negative. Check the answer: with \(x = 7\) the bed is \(15\) m by \(4\) m, and \(15 \times 4 = 60\) m\(^2\) as stated. The question asks for the length, not for \(x\), so substituting back into \((2x + 1)\) is the step that earns the mark.