TMUA: Quadratics and the discriminant
Quadratics are the most common single topic. The discriminant decides how many real roots there are, and completing the square gives the vertex.
- Section 1 Part 1 · MM1 Algebra and functions · spec MM1.3
- Papers 1 and 2
- 7 practice questions
- No calculator
What the specification covers
- Completing the square and reading off the vertex
- Discriminant: number of real roots, conditions on a parameter
- Quadratics in disguise (in x², in 2^x, in sin x)
Key ideas
- \(ax^2+bx+c=0\) has two, one or no real roots as \(b^2-4ac\) is positive, zero or negative.
- \(x^2+bx+c=\left(x+\tfrac b2\right)^2+c-\tfrac{b^2}{4}\): the vertex is \(\left(-\tfrac b2,\;c-\tfrac{b^2}{4}\right)\).
- For roots \(\alpha,\beta\) of \(ax^2+bx+c\): \(\alpha+\beta=-\tfrac ba\) and \(\alpha\beta=\tfrac ca\). Then \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\).
- A condition on a parameter \(k\) often becomes a quadratic inequality in \(k\): sketch it and read off where it is negative.
Common mistakes
- 'No real roots' is \(b^2-4ac<0\), strict; 'real roots' includes the repeated case \(\ge 0\).
- Check the coefficient of \(x^2\) is not zero before using the discriminant.
Exam tip
Use sum and product of roots whenever a question asks about the roots but not the roots themselves.
Worked example
Worked example
The equation \(x^2+(k-2)x+2k-7=0\) has no real roots. Which of the following describes all possible values of \(k\)?
- \(4 < k < 8\)
- \(k < -8\) or \(k > -4\)
- \(-8 < k < -4\)
- \(2 < k < 16\)
- \(k < 4\) or \(k > 8\)
Answer: A: \(4 < k < 8\)
No real roots means the discriminant is negative: \((k-2)^2-4(2k-7)<0\), i.e. \(k^2-12k+32<0\), i.e. \((k-4)(k-8)<0\). So \(4
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
The equation \(x^2-2kx+(-3k+1)=0\) has a repeated root for exactly two values of the constant \(k\). What is the sum of the squares of these two values of \(k\)?
- \(9\)
- \(10\)
- \(11\)
- \(7\)
- \(2\)
Show the answer and solution
Answer: C: \(11\)
A repeated root needs discriminant zero: \(4k^2-4(-3k+1)=0\), i.e. \(k^2+3k-1=0\). If the roots are \(k_1,k_2\) then \(k_1+k_2=-3\) and \(k_1k_2=-1\), so \(k_1^2+k_2^2=(k_1+k_2)^2-2k_1k_2=9+2=11\).
Question 2
The equation \(x^2-2kx+(-3k+8)=0\) has a repeated root for exactly two values of the constant \(k\). What is the sum of the squares of these two values of \(k\)?
- \(9\)
- \(17\)
- \(25\)
- \(-7\)
- \(16\)
Show the answer and solution
Answer: C: \(25\)
A repeated root needs discriminant zero: \(4k^2-4(-3k+8)=0\), i.e. \(k^2+3k-8=0\). If the roots are \(k_1,k_2\) then \(k_1+k_2=-3\) and \(k_1k_2=-8\), so \(k_1^2+k_2^2=(k_1+k_2)^2-2k_1k_2=9+16=25\).
Question 3
The equation \(x^2-2kx+(3k+8)=0\) has a repeated root for exactly two values of the constant \(k\). What is the sum of the squares of these two values of \(k\)?
- \(-7\)
- \(16\)
- \(25\)
- \(17\)
- \(9\)
Show the answer and solution
Answer: C: \(25\)
A repeated root needs discriminant zero: \(4k^2-4(3k+8)=0\), i.e. \(k^2-3k-8=0\). If the roots are \(k_1,k_2\) then \(k_1+k_2=3\) and \(k_1k_2=-8\), so \(k_1^2+k_2^2=(k_1+k_2)^2-2k_1k_2=9+16=25\).
Question 4
The equation \(x^2-2kx+(2k+5)=0\) has a repeated root for exactly two values of the constant \(k\). What is the sum of the squares of these two values of \(k\)?
- \(10\)
- \(14\)
- \(4\)
- \(-6\)
- \(9\)
Show the answer and solution
Answer: B: \(14\)
A repeated root needs discriminant zero: \(4k^2-4(2k+5)=0\), i.e. \(k^2-2k-5=0\). If the roots are \(k_1,k_2\) then \(k_1+k_2=2\) and \(k_1k_2=-5\), so \(k_1^2+k_2^2=(k_1+k_2)^2-2k_1k_2=4+10=14\).
Question 5
The equation \(x^2-2kx+(5k+8)=0\) has a repeated root for exactly two values of the constant \(k\). What is the sum of the squares of these two values of \(k\)?
- \(16\)
- \(9\)
- \(33\)
- \(25\)
- \(41\)
Show the answer and solution
Answer: E: \(41\)
A repeated root needs discriminant zero: \(4k^2-4(5k+8)=0\), i.e. \(k^2-5k-8=0\). If the roots are \(k_1,k_2\) then \(k_1+k_2=5\) and \(k_1k_2=-8\), so \(k_1^2+k_2^2=(k_1+k_2)^2-2k_1k_2=25+16=41\).
Question 6
The equation \(x^2-2kx+(3k+4)=0\) has a repeated root for exactly two values of the constant \(k\). What is the sum of the squares of these two values of \(k\)?
- \(9\)
- \(13\)
- \(1\)
- \(8\)
- \(17\)
Show the answer and solution
Answer: E: \(17\)
A repeated root needs discriminant zero: \(4k^2-4(3k+4)=0\), i.e. \(k^2-3k-4=0\). If the roots are \(k_1,k_2\) then \(k_1+k_2=3\) and \(k_1k_2=-4\), so \(k_1^2+k_2^2=(k_1+k_2)^2-2k_1k_2=9+8=17\).
Keep going
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