Question 6 - A-Level Maths

OCR C1 2007 June — Question 6 6 marks

Exam Board	OCR
Module	C1 (Core Mathematics 1)
Year	2007
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Solving quadratics and applications
Type	Substitution to solve disguised quadratic
Difficulty	Moderate -0.8 This is a straightforward substitution question requiring students to recognize that y = (x+2)² transforms the equation into y² + 5y - 6 = 0, solve this simple quadratic by factoring, then substitute back. It's more routine than average since the substitution is explicitly given and the resulting quadratic factors easily, requiring only basic algebraic manipulation.
Spec	1.02f Solve quadratic equations: including in a function of unknown

6 By using the substitution $y = ( x + 2 ) ^ { 2 }$, find the real roots of the equation $$( x + 2 ) ^ { 4 } + 5 ( x + 2 ) ^ { 2 } - 6 = 0$$

Show mark scheme Show mark scheme source

Question 6:

Answer	Marks	Guidance
Let $y = (x+2)^2$, $y^2 + 5y - 6 = 0$	B1	Substitute for $(x+2)^2$ to get $y^2 + 5y - 6 = 0$
$(y+6)(y-1) = 0$	M1, A1	Correct method to find roots; both values for $y$ correct
$y = -6$ or $y = 1$
$(x+2)^2 = 1$	M1	Attempt to work out $x$
$x = -1$ or $x = -3$	A1, A1 (6)	One correct value; second correct value and no extra real values

# Question 6:

Let $y = (x+2)^2$, $y^2 + 5y - 6 = 0$ | B1 | Substitute for $(x+2)^2$ to get $y^2 + 5y - 6 = 0$
$(y+6)(y-1) = 0$ | M1, A1 | Correct method to find roots; both values for $y$ correct
$y = -6$ or $y = 1$ | |
$(x+2)^2 = 1$ | M1 | Attempt to work out $x$
$x = -1$ or $x = -3$ | A1, A1 (6) | One correct value; second correct value and no extra real values

---

Show LaTeX source

6 By using the substitution $y = ( x + 2 ) ^ { 2 }$, find the real roots of the equation

$$( x + 2 ) ^ { 4 } + 5 ( x + 2 ) ^ { 2 } - 6 = 0$$

\hfill \mbox{\textit{OCR C1 2007 Q6 [6]}}

This paper (10 questions)

View full paper

Q1 3 Q2 5 Q3 5 Q4 5 Q5 6 Q6 6 Q7 9 Q8 9 Q9 12 Q10 12

Answer	Marks	Guidance
Let \(y = (x+2)^2\), \(y^2 + 5y - 6 = 0\)	B1	Substitute for \((x+2)^2\) to get \(y^2 + 5y - 6 = 0\)
\((y+6)(y-1) = 0\)	M1, A1	Correct method to find roots; both values for \(y\) correct
\(y = -6\) or \(y = 1\)
\((x+2)^2 = 1\)	M1	Attempt to work out \(x\)
\(x = -1\) or \(x = -3\)	A1, A1 (6)	One correct value; second correct value and no extra real values