Unit 7
5 marks · Normala) Convert the recurring decimal into a fraction, giving your answer in its simplest form.
b) The value you obtain in (a) is used as a coefficient in the quadratic equation . Solve this equation for , giving your answers in terms of .
Convert into a fraction in its simplest form.
Let be your answer to (a). Solve in terms of . Give your answer for in terms of .
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) 13/99
(b) x = [1 \pm \sqrt{1 - 8k^2}]/(2k)
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Let x = 0.1\dot{3}\dot{7} and correctly set up the equation for converting the recurring decimal.
A1 (a) · 1 mark · Correct fraction in simplest form (13/99).
M2 (b) · 1 mark · Correct substitution of k into the quadratic equation and attempt to solve for x in terms of k with the quadratic formula. (follow-through)
M3 (b) · 1 mark · States or uses the quadratic formula correctly. (follow-through)
A2 (b) · 1 mark · Accurate final answer for x in terms of k. (follow-through)
Worked solution
a) Let .
Recognize that 0.1373737… has two recurring digits: "37".
But only the '37' repeats after the tenths, so we must work with . To do this:
But 13.6 also has the non-repeating "1", so we use:
Let y = 0.037373... (the recurring part).
Make common denominators:
Alternatively, treat full period: Multiply x by 100, 100x = 13.737373..., subtract x = 0.137373...:
Check division: (from recurring decimal of "13", period two). But since "13" is the recurring part, the answer is .
Final answer: .
b) Let . The quadratic equation is .
To solve, use the quadratic formula:
Substitute if required, or leave in terms of .