Question 5:
It is given that f(x) = x2 + 3x – 4.
(a) Find the range of values of x such that f(x) > 0. [2 marks]
(b) The equation px + q – f(x) = 0, such that p and q are constants, has roots α and β.
(i) State α + β and αβ in terms of p and/or q.
(ii) The roots of the quadratic equation x2 + rx + r = 10 are 2/α and 2/β, such that r is a constant.
Express p in terms of q. Express p in terms of q.
[5 marks]
Answer:
(a) $$ \begin{aligned} f(x) & >0 \\ x^2+3 x-4 & >0 \\ (x+4)(x-1) & >0 \end{aligned} $$
$$ x<-4, x>1 $$
(b)(i) $$ \begin{aligned} & p x+q-\left(x^2+3 x-4\right)=0 \\ & -x^2+(p-3) x+q+4=0 \\ & x^2-(p-3) x-q-4=0 \\ & \alpha+\beta=p-3 \\ & \alpha \beta=-q-4 \end{aligned} $$
(b)(ii) $$ x^2+r x+r-10=0 $$
$$ \begin{aligned} \frac{2}{\alpha}+\frac{2}{\beta} & =-r \text { and }\left(\frac{2}{\alpha}\right)\left(\frac{2}{\beta}\right)=r-10 \\ \frac{2}{\alpha}+\frac{2}{\beta} & =-r \\ 2\left(\frac{\beta+\alpha}{\alpha \beta}\right) & =-r \\ \frac{2(p-3)}{-q-4} & =-r \\ r & =\frac{2(p-3)}{q+4} \end{aligned} $$
$$ \begin{aligned} \left(\frac{2}{\alpha}\right)\left(\frac{2}{\beta}\right) & =r-10 \\ \frac{4}{\alpha \beta} & =r-10 \\ \frac{4}{-q-4} & =r-10 \\ r & =\frac{-4}{q+4}+10 \end{aligned} $$
$$ \begin{aligned} \frac{2(p-3)}{q+4} & =\frac{-4}{q+4}+10 \\ 2(p-3) & =-4+10(q+4) \\ 2(p-3) & =10 q+36 \\ p-3 & =5 q+18 \\ p & =5 q+21 \end{aligned} $$
It is given that f(x) = x2 + 3x – 4.
(a) Find the range of values of x such that f(x) > 0. [2 marks]
(b) The equation px + q – f(x) = 0, such that p and q are constants, has roots α and β.
(i) State α + β and αβ in terms of p and/or q.
(ii) The roots of the quadratic equation x2 + rx + r = 10 are 2/α and 2/β, such that r is a constant.
Express p in terms of q. Express p in terms of q.
[5 marks]
Answer:
(a) $$ \begin{aligned} f(x) & >0 \\ x^2+3 x-4 & >0 \\ (x+4)(x-1) & >0 \end{aligned} $$
$$ x<-4, x>1 $$
(b)(i) $$ \begin{aligned} & p x+q-\left(x^2+3 x-4\right)=0 \\ & -x^2+(p-3) x+q+4=0 \\ & x^2-(p-3) x-q-4=0 \\ & \alpha+\beta=p-3 \\ & \alpha \beta=-q-4 \end{aligned} $$
(b)(ii) $$ x^2+r x+r-10=0 $$
$$ \begin{aligned} \frac{2}{\alpha}+\frac{2}{\beta} & =-r \text { and }\left(\frac{2}{\alpha}\right)\left(\frac{2}{\beta}\right)=r-10 \\ \frac{2}{\alpha}+\frac{2}{\beta} & =-r \\ 2\left(\frac{\beta+\alpha}{\alpha \beta}\right) & =-r \\ \frac{2(p-3)}{-q-4} & =-r \\ r & =\frac{2(p-3)}{q+4} \end{aligned} $$
$$ \begin{aligned} \left(\frac{2}{\alpha}\right)\left(\frac{2}{\beta}\right) & =r-10 \\ \frac{4}{\alpha \beta} & =r-10 \\ \frac{4}{-q-4} & =r-10 \\ r & =\frac{-4}{q+4}+10 \end{aligned} $$
$$ \begin{aligned} \frac{2(p-3)}{q+4} & =\frac{-4}{q+4}+10 \\ 2(p-3) & =-4+10(q+4) \\ 2(p-3) & =10 q+36 \\ p-3 & =5 q+18 \\ p & =5 q+21 \end{aligned} $$