for this quadratic equation:
y2-4y-16 = 0
which two options are correct solutions, rounded to three decimal places?
1) -4.483
2) -3.243
3)-2.472
4) -1.472
5) -0.536
6) 5.243
7) 6.472
8) 12.443
You have to use the quadratic equation, we will be happy to check your work.
the only solution that i get which equals to 0 is :
7) 6.472 is this correct??
i cant seem to find the other solution
To find the solutions for the quadratic equation, you can use the quadratic formula. The quadratic formula states that for an equation in the form of ax^2 + bx + c = 0, where a, b, and c are constants, the solutions x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In your case, the equation is y^2 - 4y - 16 = 0, which can be rewritten as y^2 - 4y + (-16) = 0. Here, a = 1, b = -4, and c = -16.
Let's use the quadratic formula to find the solutions:
y = (-(-4) ± √((-4)^2 - 4(1)(-16))) / (2*1)
= (4 ± √(16 + 64)) / 2
= (4 ± √80) / 2
= (4 ± 8.944) / 2
Simplifying further:
y = (4 + 8.944) / 2 = 12.944 / 2 = 6.472
y = (4 - 8.944) / 2 = -4.944 / 2 = -2.472
So, the correct solutions rounded to three decimal places are:
1) -2.472 (as stated in your question)
2) 6.472
Therefore, the correct options for the solutions in the given answer choices are 3) -2.472 and 7) 6.472.