TMUA: Simultaneous equations and inequalities
Simultaneous equations and inequalities: a line meeting a curve, and the ranges where an expression is positive.
- Section 1 Part 1 · MM1 Algebra and functions · spec MM1.4-MM1.5
- Papers 1 and 2
- 8 practice questions
- No calculator
What the specification covers
- Linear with quadratic by substitution; tangency via the discriminant
- Linear and quadratic inequalities; sketch-first method
- Never multiply an inequality by an expression of unknown sign
Key ideas
- Substitute the linear equation into the other; the resulting quadratic's discriminant says whether the line misses, touches (tangent) or cuts the curve.
- Solve a quadratic inequality by finding the roots and sketching: a positive quadratic is negative between its roots and positive outside them.
- Collect everything on one side and factorise, e.g. \((x-1)(x-4)>2(x-1)\iff(x-1)(x-6)>0\).
Common mistakes
- Never divide an inequality by an expression whose sign you do not know (such as \(x-1\)): you lose part of the solution.
- Multiplying an inequality by a negative number reverses it.
Exam tip
For inequalities, test one value from each region in the original inequality to check your answer.
Worked example
Worked example
The line \(y=mx+1\) is a tangent to the curve \(y=x^2+5x+5\). What is the sum of all possible values of \(m\)?
- \(10\)
- \(-10\)
- \(1\)
- \(8\)
- \(9\)
Answer: A: \(10\)
Substituting, \(x^2+(5-m)x+4=0\). Tangency needs a repeated root: \((5-m)^2-16=0\), so \(5-m=\pm4\), giving \(m=1\) or \(m=9\). The sum is \(10\).
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
Find the complete set of real values of \(x\) for which \((x-1)(x-4) > 2(x-1)\).
- \(x < 1\) or \(x > 4\)
- \(1 < x < 6\)
- \(4 < x < 6\)
- \(x > 6\)
- \(x < 1\) or \(x > 6\)
Show the answer and solution
Answer: E: \(x < 1\) or \(x > 6\)
Do not divide by \(x-1\): its sign is unknown. Rearranging, \((x-1)(x-4)-2(x-1)>0\), i.e. \((x-1)(x-6)>0\). A positive quadratic is positive outside its roots, so \(x<1\) or \(x>6\).
Question 2
The line \(y=mx+2\) is a tangent to the curve \(y=x^2+5x+11\). What is the product of the possible values of \(m\)?
- \(25\)
- \(10\)
- \(61\)
- \(-11\)
- \(11\)
Show the answer and solution
Answer: D: \(-11\)
Equating, \(x^2+(5-m)x+9=0\) must have a repeated root: \((5-m)^2=4\times9\), so \(m=5\pm6\). The product is \((5+6)(5-6)=25-36=-11\).
Question 3
The line \(y=mx+1\) is a tangent to the curve \(y=x^2+3x+5\). What is the product of the possible values of \(m\)?
- \(6\)
- \(25\)
- \(7\)
- \(9\)
- \(-7\)
Show the answer and solution
Answer: E: \(-7\)
Equating, \(x^2+(3-m)x+4=0\) must have a repeated root: \((3-m)^2=4\times4\), so \(m=3\pm4\). The product is \((3+4)(3-4)=9-16=-7\).
Question 4
The line \(y=mx-2\) is a tangent to the curve \(y=x^2-1x+7\). What is the product of the possible values of \(m\)?
- \(-35\)
- \(35\)
- \(-2\)
- \(1\)
- \(37\)
Show the answer and solution
Answer: A: \(-35\)
Equating, \(x^2+(-1-m)x+9=0\) must have a repeated root: \((-1-m)^2=4\times9\), so \(m=-1\pm6\). The product is \((-1+6)(-1-6)=1-36=-35\).
Question 5
The line \(y=mx+2\) is a tangent to the curve \(y=x^2+1x+11\). What is the product of the possible values of \(m\)?
- \(2\)
- \(1\)
- \(37\)
- \(-35\)
- \(35\)
Show the answer and solution
Answer: D: \(-35\)
Equating, \(x^2+(1-m)x+9=0\) must have a repeated root: \((1-m)^2=4\times9\), so \(m=1\pm6\). The product is \((1+6)(1-6)=1-36=-35\).
Question 6
The line \(y=mx+1\) is a tangent to the curve \(y=x^2+5x+2\). What is the product of the possible values of \(m\)?
- \(21\)
- \(29\)
- \(10\)
- \(-21\)
- \(25\)
Show the answer and solution
Answer: A: \(21\)
Equating, \(x^2+(5-m)x+1=0\) must have a repeated root: \((5-m)^2=4\times1\), so \(m=5\pm2\). The product is \((5+2)(5-2)=25-4=21\).
Question 7
The line \(y=mx-2\) is a tangent to the curve \(y=x^2+1x+7\). What is the product of the possible values of \(m\)?
- \(2\)
- \(37\)
- \(35\)
- \(-35\)
- \(1\)
Show the answer and solution
Answer: D: \(-35\)
Equating, \(x^2+(1-m)x+9=0\) must have a repeated root: \((1-m)^2=4\times9\), so \(m=1\pm6\). The product is \((1+6)(1-6)=1-36=-35\).
Keep going
- Previous topic: Quadratics and the discriminant
- Next topic: Polynomials, Factor and Remainder Theorems
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