Question

2y^2 - y - 8 =0 use discriminat b^2 - 4ac

Answer 1

For a second-degree polynomial of the form ay^2 + by + c = 0, we can calculate the discriminant D and get some information about the possible solutions for the equation.

If D (discriminant) is greater than 0, there will be 2 real solutions, if D is equal to 0, there will be only one real solution and if D is less than 0 there won't be a real solution.

The discriminant is given by the following formula:

D = b^2 - 4ac​

In this case, we have the equation 2y^2 - y - 8 =0, then we get:

D = (-1)^2 - 4(2)(-8) = 1 + 64 = 65

As you can see D > 0 , then the equation 2y^2 - y - 8 =0 has two different real solutions

The two solutions can be calculated by means of the quadratic formula:

`x1,x2=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

By replacing 2 for a, -1 for b and -8 for c, we get:

`\begin{gathered} x1,x2=\frac{-(-1)\pm\sqrt[]{(-1)^2-4(2)(-8)}}{2(2)}=\frac{1\pm\sqrt[]{65}}{4} \\ x1=\frac{1+\sqrt[]{65}}{4}=2.27 \\ x2=\frac{1-\sqrt[]{65}}{4}=-1.77 \end{gathered}`

Then, the two solutions to the given equation are 2.27 and -1.77