(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.
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}).