If \({\rm{x}} = \frac{{\sqrt {{\rm{a}} + {\rm{b\;}}} -{\rm{\;}}\sqrt {{\rm{a}} - {\rm{b}}} }}{{\sqrt {{\rm{a}} + {\rm{b}}} {\rm{\;}} + {\rm{\;}}\sqrt {{\rm{a}} - {\rm{b\;}}} }}\) , then what is bx 2– 2ax + b equal to (b ≠ 0)?
0
The problem asks us to find the value of the expression \({\rm{b}}{{\rm{x}}}^2 - 2{\rm{a}}{\rm{x}} + {\rm{b}}\), given a specific value for \({\rm{x}}\): \[{\rm{x}} = \frac{{\sqrt {{\rm{a}} + {\rm{b\;}}} -{\rm{\;}}\sqrt {{\rm{a}} - {\rm{b}}} }}{{\sqrt {{\rm{a}} + {\rm{b}}} {\rm{\;}} + {\rm{\;}}\sqrt {{\rm{a}} - {\rm{b\;}}} }}\] We need to perform algebraic manipulation to simplify this. A helpful technique for expressions involving ratios of square roots is the Componendo and Dividendo rule.
First, let's consider the reciprocal of \({\rm{x}}\): \[ \frac{1}{{\rm{x}}} = \frac{{\sqrt {{\rm{a}} + {\rm{b}}} {\rm{\;}} + {\rm{\;}}\sqrt {{\rm{a}} - {\rm{b\;}}} }}{{\sqrt {{\rm{a}} + {\rm{b\;}}} -{\rm{\;}}\sqrt {{\rm{a}} - {\rm{b}}} }} \] Now, we apply the Componendo and Dividendo rule, which states that if \( \frac{p}{q} = \frac{r}{s} \), then \( \frac{p+q}{p-q} = \frac{r+s}{r-s} \). We apply this to \( \frac{1/x}{1} \): \[ \frac{{1/{\rm{x}}} + 1}{{1/{\rm{x}}} - 1} = \frac{(\sqrt {{\rm{a}} + {\rm{b}}} + \sqrt {{\rm{a}} - {\rm{b}}}) + (\sqrt {{\rm{a}} + {\rm{b}}} - \sqrt {{\rm{a}} - {\rm{b}}})}{(\sqrt {{\rm{a}} + {\rm{b}}} + \sqrt {{\rm{a}} - {\rm{b}}}) - (\sqrt {{\rm{a}} + {\rm{b}}} - \sqrt {{\rm{a}} - {\rm{b}}})} \] Simplifying the numerator and the denominator: Numerator: \( (\sqrt {{\rm{a}} + {\rm{b}}} + \sqrt {{\rm{a}} - {\rm{b}}}) + (\sqrt {{\rm{a}} + {\rm{b}}} - \sqrt {{\rm{a}} - {\rm{b}}}) = 2\sqrt {{\rm{a}} + {\rm{b}}} \) Denominator: \( (\sqrt {{\rm{a}} + {\rm{b}}} + \sqrt {{\rm{a}} - {\rm{b}}}) - (\sqrt {{\rm{a}} + {\rm{b}}} - \sqrt {{\rm{a}} - {\rm{b}}}) = 2\sqrt {{\rm{a}} - {\rm{b}}} \) So, the equation becomes: \[ \frac{\frac{1}{{\rm{x}}} + 1}{\frac{1}{{\rm{x}}} - 1} = \frac{2\sqrt {{\rm{a}} + {\rm{b}}}}{2\sqrt {{\rm{a}} - {\rm{b}}}} = \sqrt{\frac{{\rm{a}} + {\rm{b}}}{{\rm{a}} - {\rm{b}}}} \] We can rewrite the left side as \( \frac{1+x}{1-x} \). Therefore: \[ \frac{1+x}{1-x} = \sqrt{\frac{{\rm{a}} + {\rm{b}}}{{\rm{a}} - {\rm{b}}}} \] Squaring both sides to eliminate the square root: \[ \left(\frac{1+x}{1-x}\right)^2 = \frac{{\rm{a}} + {\rm{b}}}{{\rm{a}} - {\rm{b}}} \] \[ \frac{(1+x)^2}{(1-x)^2} = \frac{{\rm{a}} + {\rm{b}}}{{\rm{a}} - {\rm{b}}} \] \[ \frac{1 + 2x + x^2}{1 - 2x + x^2} = \frac{{\rm{a}} + {\rm{b}}}{{\rm{a}} - {\rm{b}}} \] Now, we apply Componendo and Dividendo again on this equation: \[ \frac{(1 + 2x + x^2) + (1 - 2x + x^2)}{(1 + 2x + x^2) - (1 - 2x + x^2)} = \frac{({\rm{a}} + {\rm{b}}) + ({\rm{a}} - {\rm{b}})}{({\rm{a}} + {\rm{b}}) - ({\rm{a}} - {\rm{b}})} \] Simplify the numerator and denominator on both sides: Left side numerator: \( (1 + 2x + x^2) + (1 - 2x + x^2) = 2 + 2x^2 \) Left side denominator: \( (1 + 2x + x^2) - (1 - 2x + x^2) = 4x \) Right side numerator: \( ({\rm{a}} + {\rm{b}}) + ({\rm{a}} - {\rm{b}}) = 2{\rm{a}} \) Right side denominator: \( ({\rm{a}} + {\rm{b}}) - ({\rm{a}} - {\rm{b}}) = 2{\rm{b}} \) The equation becomes: \[ \frac{2 + 2x^2}{4x} = \frac{2{\rm{a}}}{2{\rm{b}}} \] \[ \frac{2(1 + x^2)}{4x} = \frac{\rm{a}}{\rm{b}} \] \[ \frac{1 + x^2}{2x} = \frac{\rm{a}}{\rm{b}} \] Cross-multiplying gives: \[ {\rm{b}}(1 + x^2) = 2{\rm{a}}x \] \[ {\rm{b}} + {\rm{b}}{{\rm{x}}}^2 = 2{\rm{a}}x \]
We need to evaluate \({\rm{b}}{{\rm{x}}}^2 - 2{\rm{a}}{\rm{x}} + {\rm{b}}\). From the previous step, we derived the relationship:
\[ {\rm{b}} + {\rm{b}}{{\rm{x}}}^2 = 2{\rm{a}}x \]To find the value of \({\rm{b}}{{\rm{x}}}^2 - 2{\rm{a}}{\rm{x}} + {\rm{b}}\), we can rearrange the derived equation. Subtract \( 2{\rm{a}}x \) from both sides:
\[ {\rm{b}}{{\rm{x}}}^2 - 2{\rm{a}}x + {\rm{b}} = 0 \]Thus, the value of the expression \({\rm{b}}{{\rm{x}}}^2 - 2{\rm{a}}{\rm{x}} + {\rm{b}}\) is 0.
The simplification process using algebraic manipulation, specifically the Componendo and Dividendo rule, leads directly to the relationship \({\rm{b}}{{\rm{x}}}^2 - 2{\rm{a}}{\rm{x}} + {\rm{b}} = 0\). This confirms that the value of the given expression is 0.
The correct option is the one that equals 0.
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