Question

a) Solve the following equation and calculate the answer correct to two decimal places:
$x^2-5x-10=0$.

b) Without solving the following quadratic equation, find the value of 'p' for which the given equation has real and equal roots:
$x^2+(p-3)x+p=0$.

[6 MARKS]

Answer 1

Each part: 3 Marks

a) We know that the roots of a quadratic equation are given by:
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{5\pm \sqrt{(5{)}^2-4\times 1\times (-10)}}{2\times 1}=\frac{5\pm \sqrt{25+40}}{2}=\frac{5\pm \sqrt{65}}{2}=\frac{5\pm 8.062}{2}Eitherx=\frac{13.062}{2}orx=\frac{-3.062}{2}\Rightarrow x=6.531orx=-1.531\Rightarrow x=6.53orx=-1.53$.

b)
For real and equal roots:
$b^2-4ac=0\Rightarrow (p-3{)}^2-4(1\left)\right(p)=0\Rightarrow p^2-6p+9-4p=0\Rightarrow p^2-10p+9=0\Rightarrow p^2-9p-p+9=0\Rightarrow (p-9\left)\right(p-1)=0\Rightarrow p=9orp=1$.