4y(y-3) + 5 = 12(y-3)

Question

step by step solve equation

Answer 1

ax² + bx + c = 0
D = b² - 4ac
There are two real solutions, if D > 0
There is one real solution, if D = 0
There are no real solution, if D < 0
x₁₂ = (- b ± √D) / 2a
~~~~~~~~
4y(y - 3) + 5 = 12(y - 3)
4y² - 12y + 5 - 12y + 36 = 0
4y² - 24y + 41 = 0
D = (- 24)² - 4 • 4 • 41 = - 80 < 0
Given equation does not have real solutions.

 x₁₂ = [- (-24) ± √(- 80)] / (2 • 4)

x₁ = (24 + 4 i √5) / 8 = (6 + i √5) / 2

x₂ = (24 - 4 i √5) / 8 = (6 - i √5) / 2  

Answer 2

First multiply 4y times the whole equation y-3.  You get 4y²-12y+5 = 12(y-3).  Then take the 2nd side 12(y-3) and multiply everything in the parenthesis by 12 like you did with the left side.  You get 12y-36.  The equation comes to

4y²-12y+5=12y-36.  Now you want to set one side equal to 0.  I typically take the side with the least powers to the other side by adding or subtracting each variable on both sides.  

4y²-12y+5-12y+36= 12y-36-12y+36.  After you subtracted 12y and added 36 from both sides, the right side is equal to 0 and you just solve the left side:

4y²-24y+41=0

Now first you see if you can factor this equation and solve it that way.  This looks like an equation that can not be factored so you will use the quadratic equation.  

Using the quadratic equation: (-b + sqrt (b²-4ac))/2a and (-b - sqrt (b2-4ac))/2a

The equation you solved before has the form of ay²-by+c=0 where a = 4, b=-24, and c = 41.  Now you just plug into the quadratic equation 

24 + sqrt((-24)²-4(4)(41))/2(4)

(24 + sqrt ((-80)/8) = 3 + 1.118i

And the 2nd part of the quadratic

(-24-sqrt((-24)²-80))/8= 3-1.118i

 

This shows the solution is imaginary since we have a negative square root which is what the "i" is for.  

Answer 3

4y(y - 3) + 5 = 12(y - 3 )

we use the distributive property of multiplication:

4y² - 12y + 5 = 12y - 36

4y² -12y + 5 - 12y + 36 = 12y -12y - 36 + 36

4y² - 24y + 41 = 0

use quadratic equation:

(-b ± √(b² - 4ac))/(2a)

(24 ± √(24² - 4*4*41))/2*4

(24 ± √(576 - 656))/8

(24 ± √-80)/8

(24 ± √(-16 * 5))/8

(24 ± 4i√5)/8

3 ± (i√5)/2

 

Answer 4

First, distribute 4y and 12 into respective parentheses.

4y²-12y+5=12y-36

Move everything to the left side.

4y²-12y+5-12y+36=0

4y²-24y+41=0

y_1,2=[24±√(24²-4*4*41)]/8=[24±√-80]/8.

Since √-80 does not exist (unless you consider complex numbers!), this equation has no solution.