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(D) Given $a^{2}=2,$ we have $a=\sqrt{2}.$We need to find the value of $(a+1).$$(a+1) = \sqrt{2}+1.$To express this in the form of the given options,m

Vedclass · Accepted Answer

$\frac{a-1}{3-2a}$

Vedclass · Answer

(D) Given $a^{2}=2,$ we have $a=\sqrt{2}.$We need to find the value of $(a+1).$$(a+1) = \sqrt{2}+1.$To express this in the form of the given options,multiply and divide by $(\sqrt{2}-1)^{2}$:$(a+1) = (\sqrt{2}+1) 	imes \frac{(\sqrt{2}-1)^{2}}{(\sqrt{2}-1)^{2}}$$= \frac{[(\sqrt{2}+1)(\sqrt{2}-1)](\sqrt{2}-1)}{(\sqrt{2})^{2} + 1^{2} - 2\sqrt{2}}$$= \frac{(2-1)(\sqrt{2}-1)}{2+1-2\sqrt{2}}$$= \frac{\sqrt{2}-1}{3-2\sqrt{2}}$Since $a=\sqrt{2},$ substituting $a$ back into the expression gives $\frac{a-1}{3-2a}.$

If $a^{2}=2,$ then $(a+1)$ is equal to

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