1. **State the problem:** Find all real solutions to the equations: (i) $$16a^2 = 2\sqrt{a}$$ (ii) $$b^4 + 7b^2 - 18 = 0$$ --- 2. **Solve (i):** $$16a^2 = 2\sqrt{a}$$ - First, note that $$\sqrt{a}$$ implies $$a \geq 0$$ because the square root of a negative number is not real. - Rewrite $$\sqrt{a}$$ as $$a^{1/2}$$: $$16a^2 = 2a^{1/2}$$ - Divide both sides by 2: $$\cancel{2}8a^2 = \cancel{2}a^{1/2} \implies 8a^2 = a^{1/2}$$ - To isolate terms, divide both sides by $$a^{1/2}$$ (valid only if $$a \neq 0$$): $$\frac{8a^2}{a^{1/2}} = \frac{a^{1/2}}{a^{1/2}} \implies 8a^{2 - \frac{1}{2}} = 1$$ - Simplify the exponent: $$8a^{\frac{3}{2}} = 1$$ - Divide both sides by 8: $$a^{\frac{3}{2}} = \frac{1}{8}$$ - Rewrite $$\frac{1}{8}$$ as $$\left(\frac{1}{2}\right)^3$$: $$a^{\frac{3}{2}} = \left(\frac{1}{2}\right)^3$$ - Raise both sides to the power $$\frac{2}{3}$$ to solve for $$a$$: $$a = \left(\left(\frac{1}{2}\right)^3\right)^{\frac{2}{3}} = \left(\frac{1}{2}\right)^{3 \times \frac{2}{3}} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ - Check if $$a=0$$ is a solution by substituting back into the original equation: $$16(0)^2 = 2\sqrt{0} \implies 0 = 0$$ which is true. - Therefore, solutions for (i) are $$a=0$$ and $$a=\frac{1}{4}$$. --- 3. **Solve (ii):** $$b^4 + 7b^2 - 18 = 0$$ - Let $$x = b^2$$, then the equation becomes: $$x^2 + 7x - 18 = 0$$ - Use the quadratic formula: $$x = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times (-18)}}{2 \times 1} = \frac{-7 \pm \sqrt{49 + 72}}{2} = \frac{-7 \pm \sqrt{121}}{2}$$ - Simplify the square root: $$x = \frac{-7 \pm 11}{2}$$ - Two solutions for $$x$$: $$x_1 = \frac{-7 + 11}{2} = \frac{4}{2} = 2$$ $$x_2 = \frac{-7 - 11}{2} = \frac{-18}{2} = -9$$ - Since $$x = b^2$$ and $$b^2 \geq 0$$, discard $$x_2 = -9$$. - Solve for $$b$$: $$b^2 = 2 \implies b = \pm \sqrt{2}$$ --- **Final answers:** (i) $$a = 0, \frac{1}{4}$$ (ii) $$b = \pm \sqrt{2}$$