(z+1)(z-5)=16
how would you solve this?
multiply the left..
z^2-4z-5=16
add 9 to both sides
z^2-4z+4=25
(z-2)^2=5^2 then take the square root of each side...leading to
so z-2=5 and z-2=-5
Thanks!
To solve the equation (z + 1)(z - 5) = 16, you can follow these steps:
1. Expand the equation: Multiply the two binomials, (z + 1) and (z - 5), to obtain a quadratic equation.
(z + 1)(z - 5) = 16
z^2 - 5z + z - 5 = 16
z^2 - 4z - 5 = 16
2. Simplify the equation: Combine like terms to simplify the equation.
z^2 - 4z - 5 - 16 = 0
z^2 - 4z - 21 = 0
3. Set the equation equal to zero: Move all terms to one side of the equation to set it equal to zero.
z^2 - 4z - 21 = 0
4. Factorize or use the quadratic formula: You can factorize the quadratic equation or use the quadratic formula to find the solutions for z.
The quadratic equation, z^2 - 4z - 21 = 0, does not readily factorize, so we will use the quadratic formula:
z = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -4, and c = -21. Plug these values into the quadratic formula and solve for z:
z = (-(-4) ± √((-4)^2 - 4(1)(-21))) / (2(1))
z = (4 ± √(16 + 84)) / 2
z = (4 ± √100) / 2
z = (4 ± 10) / 2
This gives two possible solutions for z:
z₁ = (4 + 10) / 2 = 14 / 2 = 7
z₂ = (4 - 10) / 2 = -6 / 2 = -3
Therefore, the solutions to the equation (z + 1)(z - 5) = 16 are z = 7 and z = -3.