(a) Solve, correct to two decimal places, the equation \(4x^{2} = 11x + 21\). (b) A man invests £1500 for two years at compound interest. After one year, hi...

Answer Details

(a) \(4x^{2}=11x+21\Rightarrow 4x^{2}-11x-21=0\).

Using the formula with \(a=4,\ b=-11,\ c=-21\):

\[x=\frac{11\pm\sqrt{(-11)^{2}-4(4)(-21)}}{2(4)}=\frac{11\pm\sqrt{121+336}}{8}=\frac{11\pm\sqrt{457}}{8}\]

\(\sqrt{457}=21.378\). So \(x=\dfrac{11+21.378}{8}=4.047\) or \(x=\dfrac{11-21.378}{8}=-1.297\).

\(x\approx 4.05\) or \(x\approx -1.30\)

(b) Principal \(=\pounds1500\); after one year the amount is \(\pounds1560\).

(i) First-year interest \(=1560-1500=\pounds60\).

\[\text{Rate}=\frac{60}{1500}\times100\%=4\%\]

(ii) Interest for the second year is charged on \(\pounds1560\):

\[\frac{4}{100}\times1560=\pounds62.40\]

(c) Car costs \(N300{,}000\), depreciating \(25\%\) then \(20\%\).

After year 1: \(300000\times(1-0.25)=300000\times0.75=N225{,}000\).

After year 2: \(225000\times(1-0.20)=225000\times0.80=N180{,}000\).

Value after 2 years \(=N180{,}000.00\)