Chapter 1 - Equations and Inequalities - Exercise Set 1.5 - Page 160: 34

The solutions are $x=-4\pm\dfrac{3}{2}\sqrt{3}$

Work Step by Step

$(2x+8)^{2}=27$ Take the square root of both sides: $\sqrt{(2x+8)^{2}}=\pm\sqrt{27}$ $2x+8=\pm\sqrt{27}$ Substitute $27$ by $9\cdot3$ and simplify: $2x+8=\pm\sqrt{9\cdot3}$ $2x+8=\pm3\sqrt{3}$ Solve for $x$: $2x=-8\pm3\sqrt{3}$ $x=-\dfrac{8}{2}\pm\dfrac{3}{2}\sqrt{3}$ $x=-4\pm\dfrac{3}{2}\sqrt{3}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.