(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)