[tex]If \: x = \frac{ \sqrt{a + 2b} + \sqrt{a + 2b} }{ \sqrt{a + 2b} – \sqrt{a – 2b} } , \: prove \: that \: {bx}^{2} – ax + b = 0.[/tex] Only that user solve this question who know. Wrong answer will be reported.
x = $\frac{\sqrt{{a}\mathrm{{+}}{2}{b}}\mathrm{{+}}\sqrt{{a}\mathrm{{-}}{2}{b}}}{\sqrt{{a}\mathrm{{+}}{2}{b}}\mathrm{{-}}\sqrt{{a}\mathrm{{-}}{2}{b}}}$×$\frac{\sqrt{{a}\mathrm{{+}}{2}{b}}\mathrm{{+}}\sqrt{{a}\mathrm{{-}}{2}{b}}}{\sqrt{{a}\mathrm{{+}}{2}{b}}\mathrm{{+}}\sqrt{{a}\mathrm{{-}}{2}{b}}}$
Step-by-step explanation:
Question:
[tex]If \: x = \frac{ \sqrt{a + 2b} + \sqrt{a + 2b} }{ \sqrt{a + 2b} – \sqrt{a – 2b} } , \: prove \: that \: {bx}^{2} – ax + b = 0.[/tex]
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.
[tex]giving \: answer \: in \: 2 \: min[/tex]
Answer:
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