(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 \(y + x = 0\).
Question 1 Report
(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 \(y + x = 0\).
Answer Details
(a) Solve \(\sqrt{4x-3}-\sqrt{2x-5}=2\).
Isolate one surd and square:
\[\sqrt{4x-3}=2+\sqrt{2x-5}\]\[4x-3=4+4\sqrt{2x-5}+(2x-5)\]\[4x-3=2x-1+4\sqrt{2x-5}\]\[2x-2=4\sqrt{2x-5}\;\Rightarrow\;x-1=2\sqrt{2x-5}\]Square again:
\[x^{2}-2x+1=4(2x-5)=8x-20\]\[x^{2}-10x+21=0\;\Rightarrow\;(x-3)(x-7)=0\]So \(x=3\) or \(x=7\). Check both in the original equation:
- \(x=3:\ \sqrt{9}-\sqrt{1}=3-1=2\) ✓
- \(x=7:\ \sqrt{25}-\sqrt{9}=5-3=2\) ✓
Both are valid, so \(x=3\) or \(x=7\).
(b) Area enclosed by \(y^{2}=4x\) and \(y+x=0\).
The line is \(x=-y\). Substitute into the parabola:
\[y^{2}=4(-y)\;\Rightarrow\;y^{2}+4y=0\;\Rightarrow\;y(y+4)=0\]So \(y=0\) (giving \((0,0)\)) and \(y=-4\) (giving \((4,-4)\)). Integrating with respect to \(y\), the line \(x=-y\) lies to the right of the parabola \(x=\tfrac{y^{2}}{4}\) on \(-4\le y\le 0\):
\[A=\int_{-4}^{0}\left(-y-\frac{y^{2}}{4}\right)\,dy=\left[-\frac{y^{2}}{2}-\frac{y^{3}}{12}\right]_{-4}^{0}\]\[=0-\left(-\frac{16}{2}-\frac{-64}{12}\right)=0-\left(-8+\frac{16}{3}\right)=\frac{8}{3}\]The finite area is \(\dfrac{8}{3}\approx 2.67\) square units.
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