I've tried to do these problems multiple times and I'm just so confused.
1. (root x-5) + 5 = 0
2. (root 2-7x) = 2x
Thanks!
1. Subtract the 5, square both sides, and add 5.
x=30
2. Square both sides, bring over 2-7x to other side, factor
√(x-5) is always positive. So, there is no solution
√(2-7x) = 2x
2-7x = 4x^2
4x^2+7x-2 = 0
(x+2)(4x-1) = 0
so, x = -2 or x = 1/4
But since √(2-7x) is positive, x = -2 is an extraneous solution
So, x = 1/4 is the only solution
Sure, I can help you understand how to solve these problems.
1. (sqrt(x) - 5) + 5 = 0
To solve this equation, you want to isolate the variable, in this case, the sqrt(x).
Start by moving the constant term (5) to the other side of the equation by subtracting it from both sides:
(sqrt(x) - 5) = -5
Now, to eliminate the square root, square both sides of the equation:
(sqrt(x))^2 = (-5)^2
This simplifies to:
x - 5 = 25
Finally, solve for x by adding 5 to both sides of the equation:
x = 25 + 5
x = 30
So the solution to the equation is x = 30.
2. sqrt(2 - 7x) = 2x
Similar to the first problem, you want to isolate the square root term, in this case, sqrt(2 - 7x).
Start by squaring both sides of the equation:
(sqrt(2 - 7x))^2 = (2x)^2
This simplifies to:
2 - 7x = 4x^2
Rearrange the equation to have a quadratic expression equal to zero:
4x^2 + 7x - 2 = 0
Now, we can solve this quadratic equation. You can use factoring, completing the square, or the quadratic formula to find the solutions. To find the solutions using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
For our quadratic equation:
a = 4, b = 7, and c = -2
Plugging these values into the quadratic formula:
x = (-(7) ± sqrt((7)^2 - 4(4)(-2))) / (2(4))
Simplifying further:
x = (-7 ± sqrt(49 + 32)) / 8
x = (-7 ± sqrt(81)) / 8
x = (-7 ± 9) / 8
Therefore, we have two solutions:
x = (-7 + 9) / 8 = 2 / 8 = 1/4
x = (-7 - 9) / 8 = -16 / 8 = -2
So the solutions to the equation are x = 1/4 and x = -2.