How the constant term decides the roots of a quadratic
How the constant term decides the roots of a quadratic
IP 2
8 marks
The constant \(k\) is a real number.
- Show that \(x^2-6x+k=(x-3)^2+(k-9)\).[2]
- Hence explain why the equation \(x^2-6x+k=0\) has no real solutions when \(k>9\), and solve the equation when \(k=9\) and when \(k=5\).[4]
- The graph of \(y=x^2-6x+k\) has a minimum value of \(-4\). Find \(k\) and write down the \(x\)-coordinates of the points where the graph meets the \(x\)-axis.[2]
Solution:
Solution locked
680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.
Create a free account to practise similar questions.
Answer:(a) Shown (b) \((x-3)^2=9-k<0\) is impossible \(k=9\): \(x=3\) \(k=5\): \(x=1\) or \(x=5\) (c) \(k=5\); \(x=1\) and \(x=5\)
Need help? Join our IP Math tuition programme.
Learn more