How do you solve $45-sqrt(10-2x^2)=25$ ?

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Answer 1 · 2017-06-29T23:02:27

See a solution process below:

First, subtract $color(red)(45)$ from each side of the equation to isolate the radical term while keeping the equation balanced:

$-color(red)(45) + 45 - sqrt(10 - 2x^2) = -color(red)(45) + 25$

$0 - sqrt(10 - 2x^2) = -20$

$-sqrt(10 - 2x^2) = -20$

Next, square both sides of the equation to eliminate the radical while keeping the equation balanced:

$(-sqrt(10 - 2x^2))^2 = (-20)^2$

$10 - 2x^2 = 400$

Then, subtract $color(red)(10)$ from each side of the equation to isolate the $x$ term while keeping the equation balanced:

$-color(red)(10) + 10 - 2x^2 = -color(red)(10) + 400$

$0 - 2x^2 = 390$

$-2x^2 = 390$

Then, divide each side of the equation by $color(red)(-2)$ to isolate $x^2$ while keeping the equation balanced:

$(-2x^2)color(red)(-2) = 390/color(red)(-2)$

$(color(red)(cancel(color(black)(-2)))x^2)cancel(color(red)(-2)) = -195$

$x^2 = -195$

Because any number squared always produces a positive result, there is no solution for $x$ which will result in a negative 195.

Or, the solution is the null or empty set: ${O/}$

How do you solve 45-sqrt(10-2x^2)=2545-sqrt(10-2x^2)=25?