1. **State the problem:** Solve the equation $4x = \sqrt{8x + 3}$ for $x$. 2. **Formula and rules:** To solve equations involving square roots, isolate the square root term and then square both sides to eliminate the root. Remember to check for extraneous solutions after squaring. 3. **Isolate and square:** $$4x = \sqrt{8x + 3}$$ Square both sides: $$ (4x)^2 = (\sqrt{8x + 3})^2 $$ $$ 16x^2 = 8x + 3 $$ 4. **Rearrange into standard quadratic form:** $$16x^2 - 8x - 3 = 0$$ 5. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=16$, $b=-8$, $c=-3$. Calculate the discriminant: $$b^2 - 4ac = (-8)^2 - 4(16)(-3) = 64 + 192 = 256$$ 6. **Find the roots:** $$x = \frac{-(-8) \pm \sqrt{256}}{2 \times 16} = \frac{8 \pm 16}{32}$$ Two solutions: - $$x = \frac{8 + 16}{32} = \frac{24}{32} = \frac{3}{4}$$ - $$x = \frac{8 - 16}{32} = \frac{-8}{32} = -\frac{1}{4}$$ 7. **Check for extraneous solutions:** Substitute $x=\frac{3}{4}$ into original: $$4 \times \frac{3}{4} = 3$$ $$\sqrt{8 \times \frac{3}{4} + 3} = \sqrt{6 + 3} = \sqrt{9} = 3$$ Valid. Substitute $x=-\frac{1}{4}$: $$4 \times -\frac{1}{4} = -1$$ $$\sqrt{8 \times -\frac{1}{4} + 3} = \sqrt{-2 + 3} = \sqrt{1} = 1$$ Left side is $-1$, right side is $1$, not equal. So $x = -\frac{1}{4}$ is extraneous. **Final answer:** $$x = \frac{3}{4}$$