Question

Write the conjugate of the following surds and find the product of each pair:

(i) $\sqrt{3}-\sqrt{2}$

(ii) $\sqrt{5}+1$

(iii) 3 + 2$\sqrt{5}$

(iv) 5 $-\sqrt{6}$

(v) $\sqrt[4]{3}+\sqrt{27}$

(vi) $3\sqrt{2}-2\sqrt{3}$

(vii) $\sqrt{5}-2\sqrt{2}$

(viii) $\sqrt{2}-1$

(ix) $\sqrt{48}+\sqrt{18}$

(x) $\sqrt{x{y}^2}-\sqrt{x^{2}y}$

Answer 1

We know that $\left(a-b\right)\left(a+b\right)=a^2-b^2$

Note: $\sqrt{1}=1$
$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{The}\mathrm{conjugate}\mathrm{of}\sqrt{3}-\sqrt{2}\mathrm{is}\sqrt{3}+\sqrt{2} \\ \mathrm{Hence},\mathrm{product}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right) \\ ={\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2 \\ =3-2 \\ =1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{The}\mathrm{conjugate}\mathrm{of}\sqrt{5}+1\mathrm{is}\sqrt{5}-1 \\ \mathrm{Hence},\mathrm{product}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right) \\ ={\left(\sqrt{5}\right)}^2-{\left(1\right)}^2 \\ =5-1 \\ =4 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{The}\mathrm{conjugate}\mathrm{of}3+2\sqrt{5}\mathrm{is}3-2\sqrt{5} \\ \mathrm{Hence},\mathrm{product}=\left(3+2\sqrt{5}\right)\left(3-2\sqrt{5}\right) \\ ={\left(3\right)}^2-{\left(2\sqrt{5}\right)}^2 \\ =9-20 \\ =-11 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{The}\mathrm{conjugate}\mathrm{of}5-\sqrt{6}\mathrm{is}5+\sqrt{6} \\ \mathrm{Hence},\mathrm{product}=\left(5-\sqrt{6}\right)\left(5+\sqrt{6}\right) \\ ={\left(5\right)}^2-{\left(\sqrt{6}\right)}^2 \\ =25-6 \\ =19 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\mathrm{The}\mathrm{conjugate}\mathrm{of}4\sqrt{3}+\sqrt{27}is4\sqrt{3}-\sqrt{27} \\ \mathrm{Hence},\mathrm{product}=\left(4\sqrt{3}+\sqrt{27}\right)\left(4\sqrt{3}-\sqrt{27}\right) \\ ={\left(4\sqrt{3}\right)}^2-{\left(\sqrt{27}\right)}^2 \\ =48-27 \\ =21 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{vi}\right)\mathrm{The}\mathrm{conjugate}\mathrm{of}3\sqrt{2}-2\sqrt{3}\mathrm{is}3\sqrt{2}+2\sqrt{3} \\ \mathrm{Hence},\mathrm{product}=\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right) \\ ={\left(3\sqrt{2}\right)}^2-{\left(2\sqrt{3}\right)}^2 \\ =18-12 \\ =6 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vii}\right)\mathrm{The}\mathrm{conjugate}\mathrm{of}\sqrt{5}-2\sqrt{2}\mathrm{is}\sqrt{5}+2\sqrt{2} \\ \mathrm{Hence},\mathrm{product}=\left(\sqrt{5}-2\sqrt{2}\right)\left(\sqrt{5}+2\sqrt{2}\right) \\ ={\left(\sqrt{5}\right)}^2-{\left(2\sqrt{2}\right)}^2 \\ =5-8 \\ =-3 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{viii}\right)\mathrm{The}\mathrm{conjugate}\mathrm{of}\sqrt{2}-1\mathrm{is}\sqrt{2}+1 \\ \mathrm{Hence},\mathrm{product}=\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right) \\ ={\left(\sqrt{2}\right)}^2-{\left(1\right)}^2 \\ =2-1 \\ =1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ix}\right)\mathrm{The}\mathrm{conjugate}\mathrm{of}\sqrt{48}+\sqrt{18}\mathrm{is}\sqrt{48}-\sqrt{18} \\ \mathrm{Hence},\mathrm{product}=\left(\sqrt{48}+\sqrt{18}\right)\left(\sqrt{48}-\sqrt{18}\right) \\ ={\left(\sqrt{48}\right)}^2-{\left(\sqrt{18}\right)}^2 \\ =48-18 \\ =30 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{x}\right)\mathrm{The}\mathrm{conjugate}\mathrm{of}\sqrt{x{y}^2}-\sqrt{x^{2}y}\mathrm{is}\sqrt{x{y}^2}+\sqrt{x^{2}y} \\ \mathrm{Hence},\mathrm{product}=\left(\sqrt{x{y}^2}-\sqrt{x^{2}y}\right)\left(\sqrt{x{y}^2}+\sqrt{x^{2}y}\right) \\ ={\left(\sqrt{x{y}^2}\right)}^2-{\left(\sqrt{x^{2}y}\right)}^2 \\ =x{y}^2-x^{2}y \end{gathered}$$