Answer:
[tex]\textsf{(a)}\quad 3\sqrt{3}-2\sqrt{5}[/tex]
[tex]\textsf{(b)}\quad x=15\sqrt{3}-10\sqrt{5}[/tex]
Step-by-step explanation:
Given rational expression:
[tex]\dfrac{8-\sqrt{15}}{2\sqrt{3}+\sqrt{5}}[/tex]
[tex]\hrulefill[/tex]
Part (a)
To rewrite the given rational expression in the form a√3 + b√5, where a and b are integers, multiply the numerator and denominator by the conjugate of the denominator:
[tex]\dfrac{(8-\sqrt{15})(2\sqrt{3}-\sqrt{5})}{(2\sqrt{3}+\sqrt{5})(2\sqrt{3}-\sqrt{5})}[/tex]
Expand the brackets:
[tex]\dfrac{16\sqrt{3}-8\sqrt{5}-2\sqrt{15}\sqrt{3}+\sqrt{15}\sqrt{5}}{4\sqrt{3}\sqrt{3}-2\sqrt{3}\sqrt{5}+2\sqrt{3}\sqrt{5}-\sqrt{5}\sqrt{5}}[/tex]
[tex]\dfrac{16\sqrt{3}-8\sqrt{5}-2\sqrt{45}+\sqrt{75}}{4(3)-5}[/tex]
[tex]\dfrac{16\sqrt{3}-8\sqrt{5}-2\sqrt{45}+\sqrt{75}}{7}[/tex]
Rewrite 45 as the product of 3² and 5, and rewrite 75 as the product of 5² and 3:
[tex]\dfrac{16\sqrt{3}-8\sqrt{5}-2\sqrt{3^2 \cdot 5}+\sqrt{5^2\cdot 3}}{7}[/tex]
[tex]\textsf{Apply the radical rule:} \quad \sqrt{ab}=\sqrt{\vphantom{b}a}\sqrt{b}[/tex]
[tex]\dfrac{16\sqrt{3}-8\sqrt{5}-2\sqrt{3^2} \sqrt{5}+\sqrt{5^2}\sqrt{3}}{7}[/tex]
[tex]\textsf{Apply the radical rule:} \quad \sqrt{a^2}=a, \quad a \geq 0[/tex]
[tex]\dfrac{16\sqrt{3}-8\sqrt{5}-(2)3 \sqrt{5}+5\sqrt{3}}{7}[/tex]
[tex]\dfrac{16\sqrt{3}-8\sqrt{5}-6 \sqrt{5}+5\sqrt{3}}{7}[/tex]
Simplify:
[tex]\dfrac{21\sqrt{3}-14\sqrt{5}}{7}[/tex]
[tex]\dfrac{21\sqrt{3}}{7}-\dfrac{14\sqrt{5}}{7}[/tex]
[tex]3\sqrt{3}-2\sqrt{5}[/tex]
Therefore, the rewritten rational expression in the form a√3 + b√5, where a and b are integers, is:
[tex]\Large\boxed{\boxed{3\sqrt{3}-2\sqrt{5}}}[/tex]
[tex]\hrulefill[/tex]
Part (b)
Given equation:
[tex](x+5\sqrt{3})\sqrt{5}=40-2x\sqrt{3}[/tex]
To solve the given equation, begin by isolating x:
[tex]\begin{aligned}(x+5\sqrt{3})\sqrt{5}&=40-2x\sqrt{3}\\\\x\sqrt{5}+5\sqrt{3}{\sqrt{5}&=40-2x\sqrt{3}\\\\x\sqrt{5}+5\sqrt{15}&=40-2x\sqrt{3}\\\\2x\sqrt{3}+x\sqrt{5}&=40-5\sqrt{15}\\\\x\left(2\sqrt{3}+\sqrt{5}\right)&=40-5\sqrt{15}\\\\x&=\dfrac{40-5\sqrt{15}}{2\sqrt{3}+\sqrt{5}}\end{aligned}[/tex]
Factor out 5 from the numerator:
[tex]\begin{aligned}x&=\dfrac{5(8-\sqrt{15})}{2\sqrt{3}+\sqrt{5}}\\\\x&=5\cdot \dfrac{8-\sqrt{15}}{2\sqrt{3}+\sqrt{5}}\end{aligned}[/tex]
Now, substitute in the result from part (a);
[tex]\begin{aligned}x&=5\cdot (3\sqrt{3}-2\sqrt{5})\\\\x&=15\sqrt{3}-10\sqrt{5}\end{aligned}[/tex]
Therefore, the solution to the given equation is:
[tex]\Large\boxed{\boxed{x=15\sqrt{3}-10\sqrt{5}}}[/tex]
