Hello...can you please let me know if I worked this problem correctly:
10Z2^ + 8z - 7 = 16z2^ + 48z - 12
0 = 6z2^ + 30z - 5
a = 6
b = 30
c = -5
x = 30 + or - (30 2^ -4(6)(-5) / 12
x = -30 + 31.937 / 12 = .161
x = -30 - 31.937 / 12 -5.161
Am I on the right track?
Thank you
right track, but you made a mistake right at the start:
6z^2 + 40z - 5 = 0
z = (-40±√(40^2+4*6*5))/12
= (-40±√(1600-120))/12
= (-40±√1480)/12
= (-40±2√370)/12
= (-20±√370)/6
z = -0.127 or -6.539
To solve the equation 6z^2 + 30z - 5 = 0, you are on the right track but your calculation for x is incorrect.
To solve a quadratic equation, you can use the quadratic formula which is given as:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 6, b = 30, and c = -5. So substituting those values into the quadratic formula, we have:
x = (-(30) ± √((30)^2 - 4(6)(-5)) / (2(6))
x = (-30 ± √(900 + 120)) / 12
x = (-30 ± √1020) / 12
Now, let's simplify the expression inside the square root:
√1020 = √(4 * 255) = 2√255
Plugging this back into the quadratic formula:
x = (-30 ± 2√255) / 12
Now, we can simplify further by factoring out a common factor of 2 in the numerator:
x = (2(-15 ± √255)) / 12
x = (-15 ± √255) / 6
So, the two solutions to the equation 6z^2 + 30z - 5 = 0 are:
x = (-15 + √255) / 6
x = (-15 - √255) / 6
Therefore, your calculations for x are not correct. The correct solutions are (-15 + √255) / 6 and (-15 - √255) / 6.