(a) Solve the equation : (sqrt{4x – 3} – sqrt{2x – 5} = 2).
(b) Find the finite area enclosed by the curve (y^{2} = 4x) and the line (y + x = 0).
Explanation
(a) (sqrt{4x – 3} – sqrt{2x – 5} = 2)
Squaring both sides, we have
(4x – 3 – 2sqrt{8x^{2} – 26x + 15} + 2x – 5 = 4)
(6x – 8 – 4 = 2sqrt{8x^{2} – 26x + 15})
(6x – 12 = 2sqrt{8x^{2} – 26x + 15})
(3x – 6 = sqrt{8x^{2} – 26x + 15})
Squaring both sides, we have
(9x^{2} – 36x + 36 = 8x^{2} – 26x + 15)
(9x^{2} – 8x^{2} – 36x + 26x + 36 – 15 = 0)
(x^{2} – 10x + 21 = 0)
Factorising, we have
((x – 7)(x – 3) = 0)
(text{x = 7 or 3}).
(b)
We integrate with respect to y.
(y^{2} = 4x implies x = frac{y^{2}}{4})
(x + y = 0 implies x = -y)
To get the boundary, equate the two values of x;
(frac{y^{2}}{4} = – y implies y^{2} = -4y)
(y^{2} + 4y = 0 implies y(y + 4) = 0)
(y = 0; y = -4)
(int_{-4}^{0} (frac{y^{2}}{4}) mathrm {d} y = [frac{y^{3}}{12}]|_{-4}^{0})
(0 – frac{-4^{3}}{12} = 0 – frac{-64}{12} = frac{64}{12} sq. units)
(int_{-4}^{0} – y mathrm {d} y = [-frac{y^{2}}{2}]|_{-4}^{0})
(0 – frac{-(-4^{2})}{2} = frac{16}{2} = 8 sq. units)
Area enclosed = (8 – frac{64}{12} = frac{32}{12} = 2frac{2}{3} sq. units)