A quadratic equation has solutions \(x=p\) and \(x=-2p\), where \(p\) is a positive integer. Write the equation in the form \(x^2+bx+c=0\), where \(b\) and \(c\) are expressed in terms of \(p\).[2]
A quadratic curve intersects the \(x\)-axis at \(x=2+\sqrt a\) and \(2-\sqrt a\). Find the equation of its line of symmetry.[1]
A straight line passes through \(P(-5,9)\) and \(Q(-9,3)\). Find the column vector \(\overrightarrow{PQ}\).[2]
Write down another column vector that is parallel to \(\overrightarrow{PQ}\).[1]
Solution:
Solution locked
Sign in to view the step-by-step solution
Similar questions are unavailable for this question.
Answer:(a) \(x^2+px-2p^2=0\) (b) \(x=2\) (c)(i) \(\begin{pmatrix}-4\\-6\end{pmatrix}\) (ii) e.g. \(\begin{pmatrix}2\\3\end{pmatrix}\)