Question

How do you simplify surds in brackets?

Answer 1

Hint: We first explain the meaning of the surds and irrational numbers. We multiply the terms according to their multiplication and the numbers inside the root values. There are four multiplications to be done. We complete all four multiplications according to the previously mentioned process. We then try to solve the surd values into their simplest form if possible.

Complete step-by-step solution:
We take multiplication of two unequal surds. We have to do the breakings of the surds in order of FOIL. The word FOIL stands for First-Outside-Inside-Last. It is a technique to distribute the multiplication of any polynomials.
We take two surds $\left( a+\sqrt{b} \right)$ and $\left( c+ \sqrt{d} \right)$.
There are two terms in each polynomial.
We start by multiplying the first terms of $\left( a+\sqrt{b} \right)$ and $\left( c+ \sqrt{d} \right)$. The terms are $a$ and $c$.
The multiplication gives the result of $a\times c=ac$.
We now multiply the outside terms of $\left( a+\sqrt{b} \right)$ and $\left( c+ \sqrt{d} \right)$. The terms are $a$ and $\sqrt{d}$.
The multiplication gives the result of $a\times \sqrt{d}=a\sqrt{d}$.
Then we multiply the inside terms of $\left( a+\sqrt{b} \right)$ and $\left( c+ \sqrt{d} \right)$. The terms are $\sqrt{b}$ and $c$.
The multiplication gives the result of $\sqrt{b}\times c=c\sqrt{b}$.
We end by multiplying the last terms of $\left( a+\sqrt{b} \right)$ and $\left( c+\sqrt{d} \right)$. The terms are $\sqrt{b}$ and $\sqrt{d}$.
The multiplication gives the result of $\sqrt{b}\times \sqrt{d}=\sqrt{bd}$.
Now we add all the terms and get the final solution as
$\left( a+\sqrt{b} \right)\left( c+\sqrt{d} \right)=ac+a\sqrt{d}+c\sqrt{b}+\sqrt{bd}$.
Therefore, multiplied value of $\left( a+\sqrt{b} \right)\left( c+\sqrt{d} \right)$ is $ac+a\sqrt{d}+c\sqrt{b}+\sqrt{bd}$.

Note: We can also simplify the irrational values if in any cases we can find the prime factors inside the root in pair form. for example: if we are taking multiplication of $\sqrt{bd}$ where $b=5,d=15$. In the multiplication we can have 5 in pair form as $15=3\times 5$.
So, $\sqrt{bd}=\sqrt{5\times 15}=\sqrt{5\times 3\times 5}=5\sqrt{3}$.