Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{4}{x}-3=\frac{5}{2x+3}, \qquad x\ne 0,-\frac{3}{2} \]Solution
Given:
\[ \frac{4}{x}-3=\frac{5}{2x+3} \]Multiplying both sides by \(x(2x+3)\):
\[ 4(2x+3)-3x(2x+3)=5x \] \[ 8x+12-6x^2-9x=5x \] \[ -6x^2-6x+12=0 \] \[ x^2+x-2=0 \]Factorizing:
\[ x^2+2x-x-2=0 \] \[ x(x+2)-1(x+2)=0 \] \[ (x+2)(x-1)=0 \]Therefore,
\[ x+2=0 \quad \text{or} \quad x-1=0 \] \[ x=-2 \quad \text{or} \quad x=1 \]Both values satisfy the condition \(x\ne 0,-\frac{3}{2}\).