Question 1 Report
(a) Given the expression \(y = ax^{2} - bx - 12\) , find the values of x when a = 1, b = 2 and y = 3.
(b) If \(\sqrt{x^{2} + 1} = \frac{5}{4}\), find the positive value of x.
(a) Substitute \(a=1,\ b=2,\ y=3\) into \(y = ax^2 - bx - 12\):
\[3 = x^2 - 2x - 12 \Rightarrow x^2 - 2x - 15 = 0.\]
Factorise: \((x - 5)(x + 3) = 0\), giving
\[x = 5\quad\text{or}\quad x = -3.\]
(b) Square both sides of \(\sqrt{x^2 + 1} = \tfrac{5}{4}\):
\[x^2 + 1 = \frac{25}{16} \Rightarrow x^2 = \frac{25}{16} - 1 = \frac{9}{16}.\]
\[x = \frac{3}{4}\quad(\text{positive value}).\]
Answer Details
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