Question

a) Solve the following equation and give your answer correct to 3 significant figures:
$5x^2-3x-4=0$

b)
Without solving the following quadratic equation, find the value of p for which the roots are equal:
$p{x}^2-4x+3=0$

[6 MARKS]

Answer 1

Each subpart: 3 Marks

a) We know that the roots of a quadratic equation are given by: $\frac{-b\frac{+}{-}\sqrt{b^2-4ac}}{2a}$
On substituting the values we get:
$x=\frac{3\pm \sqrt{(-3{)}^2-4\times 5\times (-4)}}{2\times 5}=\frac{3\pm \sqrt{9+80}}{10}=\frac{3\pm 9.433}{10}Eitherx=\frac{3+9.433}{10}orx=\frac{3-9.433}{10}\Rightarrow x=\frac{12.433}{10}=1.24orx=-\frac{6.433}{10}=-0.643$.

b)
We have
$p{x}^2-4x+3=0$
Comparing it with
$a{x}^2+bx+c=0$,
we get a = p, b = -4 and c = 3.
Now for equal roots,
$b^2-4ac=0\Rightarrow (-4{)}^2-4\times p\times 3=0$
$\Rightarrow 16-12p=0\Rightarrow 12p=16$
$\Rightarrow p=\frac{16}{12}=\frac{4}{3}$.