solve 2y=7sqrt(y)-3
2y=7sqrt(y)-3
2y+3 = √y
square both sides
4y^2 + 12y + 9 = y
4y^2 + 11y + 9 = 0
y = (-11 ±√-23)/8
I am assuming you are dealing with real numbers only
since y is a complex number, and we cannot take a square root of a complex number, your equation has no real solution.
Using the substitution y=e^x,solve 2e^x=7√e^x -3
for some reason , I dropped the 7 on the right side
let's start again
square both sides
4y^2 + 12y + 9 = 49y
4y^2 - 37y + 9 = 0
y = (37 ± √1225)/8
y = 9 or y = 0.25
(it would have factored to (y-9)(4y - 1)=0 )
now now
e^x = 9
take ln of both sides
ln (e^x) = ln9
x = ln9
or
e^x = .25 or 1/4
take ln again
x = ln (1/4) = ln1 - ln4 = -ln4
x = ln9 or x = -ln4
To solve the equation 2y = 7√y - 3, we can start by isolating the square root term.
1. Start by adding 3 to both sides of the equation to isolate the square root term:
2y + 3 = 7√y
2. Square both sides of the equation to eliminate the square root:
(2y + 3)^2 = (7√y)^2
4y^2 + 12y + 9 = 49y
3. Rearrange the equation to bring it to a standard quadratic form:
4y^2 + 12y - 49y + 9 = 0
4y^2 - 37y + 9 = 0
Now we have a quadratic equation, which we can solve by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula to find the values of y.
The quadratic formula is given by:
y = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 4, b = -37, and c = 9. Plugging in these values into the formula, we get:
y = (-(-37) ± √((-37)^2 - 4(4)(9))) / (2(4))
y = (37 ± √(1369 - 144)) / 8
y = (37 ± √(1225)) / 8
y = (37 ± 35) / 8
Now we have two potential solutions for y:
1. When we use the "+" symbol in the formula:
y = (37 + 35) / 8
y = 72 / 8
y = 9
2. When we use the "-" symbol in the formula:
y = (37 - 35) / 8
y = 2 / 8
y = 1/4
So, the solutions to the equation 2y = 7√y - 3 are y = 9 and y = 1/4.