Question

Y=2x^2-7x-18. State the zeros

Answer 1

`y=2x^2-7x-18`

In this problem, we want to find the zeros for the given function above.

Unfortunately it is not factorable, so we will need to use the quadratic formula:

`x=(-b\pm√(b^2-4ac))/(2a)`

Recall that the zeros of a quadratic function are the same as the solutions or x-intercepts.

From our equation, we can get the values of a, b, and c to use in the quadratic formula:

`\begin{gathered} y=2x^2-7x-18 \\ \\ a=2,b=-7,c=-18 \end{gathered}`

When we substitute those values, we get:

`x=(-(-7)\pm√((-7)^2-4(2)(-18)))/(2(2))`

Simplify the numerator and denominator as much as possible:

`x=(7\pm√(49+144))/(4)=(7\pm√(193))/(4)`

Split this into two equations, the addition and subtraction equations:

`x=(7+√(193))/(4),\text{ and }x=(7-√(193))/(4)`

We can use a calculator to get the final values:

`\begin{gathered} x=(7+√(193))/(4)\approx5.22 \\ \\ x=(7-√(193))/(4)\approx-1.72 \end{gathered}`

Our zeros are:

`\boxed{5.22\text{ and }-1.72}`