Question

If root 7 minus 1 upon root 7 + 1 - root 7 + 1 upon root 7 minus 1 equal to a + b root 7 then find a and b

Answer 1

There is your answers.

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Answer 2

Answer:

The value of a = 0 and b = (-2/3).

Step-by-step explanation:

Given:

$\frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{7+1}{\sqrt{7}-1}=a+b \sqrt{7}$

To find:

the values of a and b

Let,

$\frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}$

LHS $=\frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}$

Cross multiplying the above equation we get

$=\frac{(\sqrt{7}-1) \times(\sqrt{7}-1)}{(\sqrt{7}+1)(\sqrt{7}-1)}-\frac{(\sqrt{7}+1) \times(\sqrt{7}+1)}{(\sqrt{7}-1) \times(\sqrt{7}+1)}$

Canceling the common terms we get

$=\frac{(\sqrt{7})^{2}+1^{2}-2 \sqrt{7}}{(\sqrt{7})^{2}-1^{2}}-\frac{(\sqrt{7})^{2}+1^{2}+2 \sqrt{7}}{(\sqrt{7})^{2}-1^{2}}$

$=\frac{8-2 \sqrt{7}}{6}-\frac{8+2 \sqrt{7}}{6}=\frac{8-2 \sqrt{7}-(8+2 \sqrt{7})}{6}$

Simplifying the above equation we get

$=\frac{8-2 \sqrt{7}-8-2 \sqrt{7}}{6}=\frac{-2 \sqrt{7}-2 \sqrt{7}}{6}$

$=\frac{\sqrt{7}(-2-2)}{6}=\frac{\sqrt{7} \times-4}{6}$

$=\frac{-4 \sqrt{7}}{63}=-\frac{2}{3} \sqrt{7}$

$0+\left(\frac{-2}{3}\right) \sqrt{7}=a+b \sqrt{7}$

Where, a = 0 then

b = (-2/3)

Therefore, the value of a = 0 and b = (-2/3).

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