Question

hey guys pls solve 14.ii,iii,iv.thanks a lot​

Answer 1

Answer:

Then general idea here is to multiply numerator and denominator by a conjugate of the denominator to get rid of the surds there.

(ii) Start by simplifying the expression for x.

$x = \frac{\sqrt{a^2+ab}+\sqrt{a^2-ab}}{\sqrt{a^2+ab}-\sqrt{a^2-ab}} \\\\ = \frac{(\sqrt{a^2+ab}+\sqrt{a^2-ab})^2}{(\sqrt{a^2+ab}-\sqrt{a^2-ab})(\sqrt{a^2+ab}+\sqrt{a^2-ab})} \\\\ = \frac{(a^2+ab)+(a^2-ab)+2\sqrt{(a^2+ab)(a^2-ab)}}{(a^2+ab)-(a^2-ab)} \\\\ = \frac{2a^2+2\sqrt{a^2(a+b)(a-b)}}{2ab} \\\\= \frac{a+\sqrt{a^2-b^2}}{b}$

From here, we then rearrange and square to get

$bx = a+\sqrt{a^2-b^2}\\\Rightarrow bx-a = \sqrt{a^2-b^2}\\\Rightarrow b^2x^2-2abx+a^2 = a^2-b^2 \\\Rightarrow b^2x^2 -2abx + b^2 = 0\\\Rightarrow bx^2 -2ax + b = 0$

(iii) Being a bit briefer this time where the steps are much the same,

$y = \frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}\\=\frac{(a+x)+(a-x)+2\sqrt{a^2-x^2}}{(a+x)-(a-x)}\\=\frac{a+\sqrt{a^2-x^2}}{x}\\\\\Rightarrow xy -a = \sqrt{a^2-x^2}\\\Rightarrow x^2y^2-2axy+a^2=a^2-x^2\\\Rightarrow xy^2-2ay+x=0\\\Rightarrow x(y^2+1) = 2ay \\\Rightarrow x=\frac{2ay}{y^2+1}$

(iv) Finally, this is just like (iii) but with 3b and x here in place of the x and the y there.  So from the working above but with the symbols changed appropriately, we get to the quadratic xy² - 2ay + x = 0 in the third last line, which this time is 3bx² - 2ax + 3b = 0.

Hope that helps!