Question

$If \: x = \frac{ \sqrt{a + 2b} + \sqrt{a + 2b} }{ \sqrt{a + 2b} - \sqrt{a - 2b} } , \: prove \: that \: {bx}^{2} - ax + b = 0.$
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Answer 1

Step-by-step explanation:

Question:

$If \: x = \frac{ \sqrt{a + 2b} + \sqrt{a + 2b} }{ \sqrt{a + 2b} - \sqrt{a - 2b} } , \: prove \: that \: {bx}^{2} - ax + b = 0.$

To prove:

that bx²-ax+b=0

Solution:

x = $\frac{\sqrt{{a}\mathrm{{+}}{2}{b}}\mathrm{{+}}\sqrt{{a}\mathrm{{-}}{2}{b}}}{\sqrt{{a}\mathrm{{+}}{2}{b}}\mathrm{{-}}\sqrt{{a}\mathrm{{-}}{2}{b}}}$×$\frac{\sqrt{{a}\mathrm{{+}}{2}{b}}\mathrm{{+}}\sqrt{{a}\mathrm{{-}}{2}{b}}}{\sqrt{{a}\mathrm{{+}}{2}{b}}\mathrm{{+}}\sqrt{{a}\mathrm{{-}}{2}{b}}}$

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

=$\frac{{a}\mathrm{{+}}{2}{b}\mathrm{{+}}{a}\mathrm{{-}}{2}{b}\mathrm{{+}}{2}\mathrm{{-}}\sqrt{{a}^{2}\mathrm{{-}}{4}{b}^{2}}}{{a}\mathrm{{+}}{2}{b}\mathrm{{-}}{\mathrm{(}}{a}\mathrm{{-}}{2}{b}{\mathrm{)}}}$

=>$\frac{{a}\mathrm{{+}}\sqrt{{a}^{2}\mathrm{{-}}{4}{b}^{2}}}{2b}$

=> $\fbox{${{2}{bx}\mathrm{{=}}{a}\mathrm{{+}}\sqrt{{a}^{2}\mathrm{{-}}{4}{b}^{2}}}$}$

=> $\fbox{${{2}{bx}\mathrm{{-}}{a}\mathrm{{=}}\sqrt{{a}^{2}\mathrm{{-}}{4}{b}^{2}}}$}$

Squaring on both sides,we get

(2bx-a²) = a²-4b²

=> 4b²x²+a²-4abx-a²+4b² = 0

=> 4b²x²-4abx+4b² = 0 => 4(b²x²-abx+b²) = 0

=> b²x²-abx+b² = 0 => b(bx²-ax+b) = 0

=> bx²-ax+b = 0 Hence, proved.

$giving \: answer \: in \: 2 \: min$

Answer 2

Answer:

see above write answer see