Solve for x.
Problem #22
25x^2 = 13
My answer: x = + - radical (13)/(5)
Problem #32
2 (x - 5 )^2 = 3
My answer: x = 5 +- radical (3)/(2)
Problem #42
Solve for x
x^2+4x+4=7 (hint factor the left-hand side)
My answer: x = -2 +- radical (7)
To solve for x in Problem #22, we have the equation 25x^2 = 13.
To begin, divide both sides of the equation by 25 to isolate x^2:
x^2 = 13/25
Next, we can take the square root of both sides of the equation to solve for x:
x = ± √(13/25)
Simplifying the square root, we get:
x = ± √13 / √25
Since √25 = 5, we can further simplify to:
x = ± √13 / 5
So the solution to Problem #22 is x = ± √13 / 5.
Now, let's solve Problem #32. We have the equation 2(x - 5)^2 = 3.
First, divide both sides of the equation by 2 to isolate (x - 5)^2:
(x - 5)^2 = 3/2
Next, take the square root of both sides:
x - 5 = ± √(3/2)
Adding 5 to both sides of the equation, we get:
x = 5 ± √(3/2)
This can be further simplified as:
x = 5 ± √3 / √2
Multiplying the numerator and the denominator of the square root by √2, we have:
x = 5 ± √(3 * 2) / 2
Simplifying further, we get:
x = 5 ± √6 / 2
Therefore, the solution to Problem #32 is x = 5 ± √6 / 2.
Moving on to Problem #42. We have the equation x^2 + 4x + 4 = 7.
To solve this equation, we can start by factoring the left-hand side:
(x + 2)(x + 2) = 7
Simplifying the factored expression, we get:
(x + 2)^2 = 7
Next, we can take the square root of both sides:
x + 2 = ± √7
Subtracting 2 from both sides gives us:
x = -2 ± √7
Therefore, the solution to Problem #42 is x = -2 ± √7.
It's important to note that when solving quadratic equations, both the plus and minus solutions are considered, unless otherwise specified by the problem or context.