Solve the following equation- -x - 3 - 4 -x - 1 - -x - 8 - 6 -x - 1 - 1

√(x + 3 4 √(x 1)) + √(x + 8 6 √(x 1)) = 1 √(x 1) = t x = t^2+1 √(t^2+4 4t)+√(t^2 9 6t) = 1 √((t 2) ...

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Byju's Answer

Standard VIII

Mathematics

Random Interesting Patterns Followed by Square Numbers

Solve the fol...

Question

Solve the following equation:
$\sqrt{x + 3 - 4 \sqrt{x - 1}} + \sqrt{x + 8 - 6 \sqrt{x - 1}} = 1$

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Solution

$\sqrt{x + 3 - 4 \sqrt{x - 1}} + \sqrt{x + 8 - 6 \sqrt{x - 1}} = 1$
$\sqrt{x - 1} = t$
$x = t^{2} + 1$
$\sqrt{t^{2} + 4 - 4 t} + \sqrt{t^{2} - 9 - 6 t} = 1$
$\sqrt{(t - 2)^{2}} + \sqrt{(t - 3)^{2}} = 1$
$t - 2 + t - 3 = 1$
$2 t = 6$ $t = 3$
$\sqrt{x - 1} = 3$
$x - 1 = 9$
$x = 10$

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Solve the following equation:√x+3−4√x−1+√x+8−6√x−1=1

√x+3−4√x−1+√x+8−6√x−1=1√x−1=tx=t2+1√t2+4−4t+√t2−9−6t=1√(t−2)2+√(t−3)2=1t−2+t−3=12t=6 t=3√x−1=3x−1=9x=10

Solve the following equation:
$\sqrt{x + 3 - 4 \sqrt{x - 1}} + \sqrt{x + 8 - 6 \sqrt{x - 1}} = 1$

$\sqrt{x + 3 - 4 \sqrt{x - 1}} + \sqrt{x + 8 - 6 \sqrt{x - 1}} = 1$
$\sqrt{x - 1} = t$
$x = t^{2} + 1$
$\sqrt{t^{2} + 4 - 4 t} + \sqrt{t^{2} - 9 - 6 t} = 1$
$\sqrt{(t - 2)^{2}} + \sqrt{(t - 3)^{2}} = 1$
$t - 2 + t - 3 = 1$
$2 t = 6$ $t = 3$
$\sqrt{x - 1} = 3$
$x - 1 = 9$
$x = 10$