(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.
(b) If \(\sqrt{x^{2} + 1} = \frac{5}{4}\), find the positive value of x.
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Correct Answer: Option n
Explanation:
(a) \(y = ax^{2} - bx - 12\)
When a = 1, b = 2 and y = 3.
\(3 = x^{2} - 2x - 12\)
\(x^{2} - 2x - 12 - 3 = 0 \implies x^{2} - 2x - 15 = 0\)
\(x^{2} - 5x + 3x - 15 = 0\)
\((x - 5)(x + 3) = 0\)
\(\text{x = 5 or -3}\)
(b) \(\sqrt{x^{2} + 1} = \frac{5}{4}\)
Squaring both sides,
\(x^{2} + 1 = \frac{25}{16}\)
\(x^{2} = \frac{25}{16} - 1 = \frac{9}{16}\)
\(x = \sqrt{\frac{9}{16}} = \pm \frac{3}{4}\)
The positive value of x = \(\frac{3}{4}\).
(a) \(y = ax^{2} - bx - 12\)
When a = 1, b = 2 and y = 3.
\(3 = x^{2} - 2x - 12\)
\(x^{2} - 2x - 12 - 3 = 0 \implies x^{2} - 2x - 15 = 0\)
\(x^{2} - 5x + 3x - 15 = 0\)
\((x - 5)(x + 3) = 0\)
\(\text{x = 5 or -3}\)
(b) \(\sqrt{x^{2} + 1} = \frac{5}{4}\)
Squaring both sides,
\(x^{2} + 1 = \frac{25}{16}\)
\(x^{2} = \frac{25}{16} - 1 = \frac{9}{16}\)
\(x = \sqrt{\frac{9}{16}} = \pm \frac{3}{4}\)
The positive value of x = \(\frac{3}{4}\).