solve the equation
1). (x+4)(x-12)=5x
2). 4x^2+2x-14=16-17x
This is an example of why we always get everything on the left side, then factor to make the product zero:
(x+4)(x-12) = 5x
x^2 - 8x - 48 = 5x
x^2 - 13x - 48 = 0
(x-16)(x+3) = 0
Now we have a zero product. That means that if any of the factors is zero, then the product is also zero.
So, x = -3 or 16
Now do the same steps with the second equation
To solve the given equations, let's follow these steps:
1) (x+4)(x-12)=5x
Expand the equation:
x^2 - 12x + 4x - 48 = 5x
Simplify:
x^2 - 8x - 48 = 5x
Rearrange all terms to one side to form a quadratic equation:
x^2 - 8x - 5x - 48 = 0
x^2 - 13x - 48 = 0
2) 4x^2+2x-14=16-17x
Rearrange the equation:
4x^2 + 17x + 2x - 14 - 16 = 0
Combine like terms:
4x^2 + 19x - 30 = 0
Now, to solve both equations, we can use factoring or the quadratic formula.
1) (x+4)(x-12)=5x
Rewrite the equation:
x^2 - 8x - 5x - 48 = 0
Factor the equation:
(x-12)(x+4) - 5(x+4) = 0
Now we have a common factor, (x+4):
(x+4)(x-12-5) = 0
(x+4)(x-17) = 0
Setting each factor equal to zero:
x + 4 = 0 or x - 17 = 0
Solving for x:
x = -4 or x = 17
Therefore, the roots for the first equation are x = -4 and x = 17.
2) 4x^2 + 19x - 30 = 0
To solve this quadratic equation, we can use factoring or the quadratic formula. Let's use the quadratic formula.
The quadratic formula is:
x = (-b ± √(b^2 - 4ac)) / 2a
In this equation:
a = 4, b = 19, c = -30
Substituting these values into the quadratic formula:
x = (-19 ± √(19^2 - 4(4)(-30))) / (2(4))
Simplifying:
x = (-19 ± √(361 + 480)) / 8
x = (-19 ± √(841)) / 8
x = (-19 ± 29) / 8
x = (-19 + 29) / 8 or x = (-19 - 29) / 8
x = 10 / 8 or x = -48 / 8
Simplifying further:
x = 5/4 or x = -6
Therefore, the roots for the second equation are x = 5/4 and x = -6.