PP1 (Shorter) Quant Section 2 (Medium) Q12

<p><span style="color:#27ae60;">Solving the Problem</span></p> <p><strong>Method 1 (Algebra)</strong></p> <p>We can start out by re-writing some of the root identities:</p> <p>$(\sqrt{4x} + \sqrt{9x})^2 = (\sqrt{4}\sqrt{x} + \sqrt{9}\sqrt{x})^2$</p> <p>This would then become</p> <ul> <li>$(\sqrt{4}\sqrt{x} + \sqrt{9}\sqrt{x})^2$</li> <li>$(2\sqrt{x} + 3\sqrt{x})^2$</li> </ul> <p>We can now expand this out:</p> <ul> <li>$(2\sqrt{x} + 3\sqrt{x})(2\sqrt{x} + 3\sqrt{x})$</li> <li>$2\sqrt{x}(2\sqrt{x})+2\sqrt{x}(3\sqrt{x})+3\sqrt{x}(2\sqrt{x})+3\sqrt{x}(3\sqrt{x})$</li> </ul> <p>And now we can simplify:</p> <p>$4x+6x+6x+9x=25x$</p> <p><strong>Method 2 (Choosing Numbers)</strong></p> <p>Since we are told that $x&gt;0$, let&#39;s choose $x=1$.</p> <p>Now, we just plug in $1$ into the equation wherever there is an x:</p> <ul> <li>$(\sqrt{4(1)} + \sqrt{9(1)})^2$</li> <li>$(\sqrt{4}+\sqrt{9})^2$</li> <li>$(|\sqrt{4}|+|\sqrt{9}|)^2$</li> <li>$(2+3)^2$</li> <li>$5^2=25$</li> </ul> <p>Now, we can see that only when we put $x=1$ into the equation $25x$, we get the result of $25$.</p> <p>So, the answer would be <span style="color:#27ae60;">D</span>.</p>