(a) Solve the equation, correct to two decimal places \(2x^{2} + 7x – 11 = 0\)
(b) Using the substitution \(P = \frac{1}{x}; Q = \frac{1}{y}\), solve the simultaneous equations : \(\frac{2}{x} + \frac{1}{y} = 3 ; \frac{1}{x} – \frac{5}{y} = 7\)
Explanation
(a) \(2x^{2} + 7x - 11 = 0\)
Using the quadratic formula,
\(a = 2, b = 7, c = -11\)
\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
\(x = \frac{-7 \pm \sqrt{7^{2} - 4(2)(-11)}}{2(2)}\)
\(x = \frac{-7 \pm \sqrt{49 + 88}}{4}\)
\(x = \frac{-7 \pm 11.705}{4}\)
\(x = \frac{-7 + 11.705}{4} ; \frac{-7 - 11.705}{4}\)
\(x = \frac{4.705}{4} = 1.17625\) or \(x = \frac{-18.705}{4} = -4.67625\)
\(x \approxeq 1.18\) or \(x \approxeq -4.68\)
(b) \(\frac{2}{x} + \frac{1}{y} = 3\)
\(\frac{1}{x} - \frac{5}{y} = 7\)
Substituting using \(P = \frac{1}{x}; Q = \frac{1}{y}\), we have
\(2P + Q = 3 ... (i)\)
\(P - 5Q = 7 ... (ii)\)
Multiply (ii) by 2, we have \(2P - 10Q = 14 ... (iii)\)
(iii) - (i) : \(-10Q - Q = 14 - 3 \implies -11Q = 11\)
\(Q = -1\)
Substitute Q = -1 in (i),
\(2P - 1 = 3 \implies 2P = 3 + 1 = 4\)
\(\implies P = 2\)
P = 2; Q = -1
But \(P = \frac{1}{x} \implies 2 = \frac{1}{x}\)
\(\therefore x = \frac{1}{2}\)
Also, \(Q = \frac{1}{y} \implies -1 = \frac{1}{y}\)
\(y = -1\)
\(\therefore x = \frac{1}{2} ; y = -1\)