Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{m}{n}x^2+\frac{n}{m}=1-2x \]Solution
Given:
\[ \frac{m}{n}x^2+\frac{n}{m}=1-2x \]Multiplying both sides by \(mn\):
\[ m^2x^2+n^2=mn-2mnx \] \[ m^2x^2+2mnx+n^2-mn=0 \]Observe that:
\[ m^2x^2+2mnx+n^2=(mx+n)^2 \] Therefore, \[ (mx+n)^2-mn=0 \]Using the identity \(a^2-b^2=(a-b)(a+b)\):
\[ (mx+n-\sqrt{mn})(mx+n+\sqrt{mn})=0 \]Hence,
\[ mx+n-\sqrt{mn}=0 \] or \[ mx+n+\sqrt{mn}=0 \]Therefore,
\[ x=\frac{\sqrt{mn}-n}{m} \] or \[ x=-\frac{n+\sqrt{mn}}{m} \]