Question

If x = √a+1 + √a−1
√a+1 − √a−1
, using properties of proportion show that x2
-2ax+1=0.

Answer 1

Answer:

The value of x is

x = 3

Given equation,

\frac{ {x}^{2} - x + 1}{ {x}^{2} + x + 1} = \frac{14(x - 1)}{13(x + 1)}x2+x+1x2−x+1=13(x+1)14(x−1)

\frac{ {x}^{2} - x + 1}{ {x}^{2} + x + 1} = \frac{14x - 14}{13x + 13} \: \: \: - (1)x2+x+1x2−x+1=13x+1314x−14−(1)

Now, using the componendo dividendo property that is

\begin{gathered}if \: \frac{a}{b} = \frac{c}{d} \\ then \: \frac{a + b}{a - b} = \frac{c + d}{c - d}\end{gathered}ifba=dcthena−ba+b=c−dc+d

We can write (1) as

\frac{ ({x}^{2} - x + 1) + ( {x}^{2} + x + 1) }{( {x}^{2} - x + 1) - ( {x}^{2} + x + 1)} = \frac{(14x - 14) + (13x + 13)}{(14x - 14) - (13x + 13)}(x2−x+1)−(x2+x+1)(x2−x+1)+(x2+x+1)=(14x−14)−(13x+13)(14x−14)+(13x+13)

\frac{ {x}^{2} - x + 1 + {x}^{2} + x + 1}{{x}^{2} - x + 1- {x}^{2} - x - 1} = \frac{14x - 14 + 13x + 13}{14x - 14 - 13x - 13}x2−x+1−x2−x−1x2−x+1+x2+x+1=14x−14−13x−1314x−14+13x+13

\frac{2 {x}^{2} + 2}{ - 2x} = \frac{27x -1}{x - 27}−2x2x2+2=x−2727x−1

(x - 27)( {x}^{2} + 1) = ( - x)(27x - 1)(x−27)(x2+1)=(−x)(27x−1)

{x}^{3} - 27 {x}^{2} + x - 27 = - 27 {x}^{2} + xx3−27x2+x−27=−27x2+x

{x}^{3} - 27 {x}^{2} + 27 {x}^{2} + x - x = 27x3−27x2+27x2+x−x=27

{x}^{3} = 27x3=27