$\displaystyle x\sqrt{{40}}=x\sqrt{5}+\sqrt{{10}}$

Bring the values of x to one side of the equation and factorize out the x.

$\displaystyle x\sqrt{{40}}-x\sqrt{5}=\sqrt{{10}}$

$\displaystyle x(\sqrt{{40}}-\sqrt{5})=\sqrt{{10}}$

$\displaystyle x=\frac{{\sqrt{{10}}}}{{(\sqrt{{40}}-\sqrt{5})}}$

Rationalize the denominator by multiplying with the conjugate of the denominator
i.e.√40 + √5

$\displaystyle x=\frac{{\sqrt{{10}}}}{{(\sqrt{{40}}-\sqrt{5})}}\times \frac{{(\sqrt{{40}}+\sqrt{5})}}{{(\sqrt{{40}}+\sqrt{5})}}$

$\displaystyle x=\frac{{\sqrt{{10}}(\sqrt{{40}}+\sqrt{5})}}{{{{{(\sqrt{{40}})}}^{2}}-{{{(\sqrt{5})}}^{2}}}}$

Simplify and change the equation into the form of a+√b ⁄ 7

$\displaystyle x=\frac{{\sqrt{{10}}(\sqrt{{40}}+\sqrt{5})}}{{40-5}}$

$\displaystyle x=\frac{{\sqrt{{400}}+\sqrt{5}0}}{{35}}$

$\displaystyle x=\frac{{20+\sqrt{{25\times 2}}}}{{35}}$

$\displaystyle x=\frac{{20+5\sqrt{2}}}{{35}}$

$\displaystyle x=\frac{{5(4+\sqrt{2})}}{{35}}$

$\displaystyle x=\frac{{4+\sqrt{2}}}{7}$

Answer: a = 4, b = 2

At first glance, student may be clueless as to how to start solving this surd problem. But if you think in terms of algebra, you will realize that you need to make x the subject of the equation before rationalizing the denominator. Do remember to multiplying with the conjugate ( means put opposite sign in the expression), a common mistake is to use the same sign, only to realize that you cannot rid yourself of the square root. Finally, you have to write out the value of a and b, don’t leave the answer as a fraction.

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