Solve. Remember to check for extraneous roots.

|2x+5| = 3x+4

$\begin{array}{rcl} |2x+5| &=& 3x+4 \qquad | \qquad ()^2\\ (2x+5)^2 &=& (3x+4)^2 \\ 4x^2+20x+25 &=& 9x^2+24x+16\\ 5x^2+4x-9 &=& 0 \\\\ \boxed{~ x = {-b \pm \sqrt{b^2-4ac} \over 2a} ~}\\ 5x^2+4x-9 &=& 0 \\ x_{1,2} &=& {-4 \pm \sqrt{16-4\cdot 5 \cdot(-9)} \over 2\cdot 5}\\ x_{1,2} &=& {-4 \pm \sqrt{16+180} \over 10}\\ x_{1,2} &=& {-4 \pm 14 \over 10}\\\\ x_1 &=& {-4 +14 \over 10}\\ x_1 &=& {10 \over 10}\\ \mathbf{x_1} &\mathbf{=}& \mathbf{1}\\ x_2 &=& {-4 -14 \over 10}\\ x_2 &=& {-18 \over 10}\\ \mathbf{x_2} &\mathbf{=}& \mathbf{-1.8} \qquad \text{no solution}\\ \end{array}$

Hallo guest

Sorry!

Hi Omi67 :)