Question

Write the conjugates of the binomial surd $3\sqrt{7}+7\sqrt{3}$

Answer 1

We know that the when sum of two terms and the difference of the same two terms are multiplied, the product is always a rational number.

Let us apply this concept to a binomial surd $(3\sqrt{7}+7\sqrt{3})$.

When we multiply this with the difference of the same two terms, that is, with $(3\sqrt{7}-7\sqrt{3})$, the product is:

$(3\sqrt{7}+7\sqrt{3})(3\sqrt{7}-7\sqrt{3})={\left(3\sqrt{7}\right)}^2-{\left(7\sqrt{3}\right)}^2=(9\times 7)-(49\times 3)=63-147=-84(\because a^2-b^2=(a+b\left)\right(a-b\left)\right)$

Since $-84$ is a rational number.

Hence, $(3\sqrt{7}-7\sqrt{3})$ is the conjugate of $(3\sqrt{7}+7\sqrt{3})$.