Substitution to solve disguised quadratic

A question is this type if and only if it requires an explicit substitution (e.g. u = x^(1/3), y = 2^x, u = (3x-2)^2) to reduce a non-standard equation into a solvable quadratic or factorable form.

8 questions · Moderate -0.4

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CAIE P1 2022 June Q5
7 marks Moderate -0.3
5
  1. Solve the equation \(6 \sqrt { y } + \frac { 2 } { \sqrt { y } } - 7 = 0\).
  2. Hence solve the equation \(6 \sqrt { \tan x } + \frac { 2 } { \sqrt { \tan x } } - 7 = 0\) for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
CAIE P1 2021 March Q2
4 marks Standard +0.3
2 By using a suitable substitution, solve the equation $$( 2 x - 3 ) ^ { 2 } - \frac { 4 } { ( 2 x - 3 ) ^ { 2 } } - 3 = 0$$
CAIE P3 2004 June Q4
6 marks Moderate -0.3
4
  1. Show that if \(y = 2 ^ { x }\), then the equation $$2 ^ { x } - 2 ^ { - x } = 1$$ can be written as a quadratic equation in \(y\).
  2. Hence solve the equation $$2 ^ { x } - 2 ^ { - x } = 1$$
OCR C1 2007 June Q6
6 marks Moderate -0.8
6 By using the substitution \(y = ( x + 2 ) ^ { 2 }\), find the real roots of the equation $$( x + 2 ) ^ { 4 } + 5 ( x + 2 ) ^ { 2 } - 6 = 0$$
OCR C1 Q6
7 marks Moderate -0.5
  1. (i) Given that \(y = x ^ { \frac { 1 } { 3 } }\), show that the equation
$$2 x ^ { \frac { 1 } { 3 } } + 3 x ^ { - \frac { 1 } { 3 } } = 7$$ can be rewritten as $$2 y ^ { 2 } - 7 y + 3 = 0 .$$ (ii) Hence, solve the equation $$2 x ^ { \frac { 1 } { 3 } } + 3 x ^ { - \frac { 1 } { 3 } } = 7$$
OCR C1 Q3
5 marks Moderate -0.3
3. (i) Solve the equation $$y ^ { 2 } + 8 = 9 y .$$ (ii) Hence solve the equation $$x ^ { 3 } + 8 = 9 x ^ { \frac { 3 } { 2 } } .$$
OCR C1 2011 January Q4
6 marks Moderate -0.5
4 By using the substitution \(u = ( 3 x - 2 ) ^ { 2 }\), find the roots of the equation $$( 3 x - 2 ) ^ { 4 } - 5 ( 3 x - 2 ) ^ { 2 } + 4 = 0$$
Edexcel AS Paper 1 2023 June Q2
4 marks Moderate -0.8
  1. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
Using the substitution \(u = \sqrt { x }\) or otherwise, solve $$6 x + 7 \sqrt { x } - 20 = 0$$