Question

How do you solve `y = -2x^2 + 5x - 1`?

Answer 1

The roots are:

`((5+sqrt17)/4,0)` and `((5-sqrt17)/4,0)`

The approximate roots are:

`(0.2192,0)` and `(2.281,0)`

Explanation:

Solve:

`y=-2x^2+5x-1` is a quadratic equation in standard form:

`y=ax^2+bx+c`,

where:

`a=-2`, `b=5`, `c=-1`

Substitute `0` for `y`.

`0=-2x^2+5x-1`

Solve using the quadratic equation.

`x=(-b+-sqrt(b^2-4ac))/(2*a)`

Plug in the known value.

`x=(-5+-sqrt(5^2-4*-2*-1))/(2*-2)`

Simplify.

`x=(-5+-sqrt(25-8))/(-4)`

`x=(-5+-sqrt17)/(-4)`

Simplify.

`x=(5+-sqrt17)/4`

`x=(5+sqrt17)/4,` `(5-sqrt17)/4`

The roots are:

`((5+sqrt17)/4,0)` and `((5-sqrt17)/4,0)`

The approximate roots are:

`(0.2192,0)` and `(2.281,0)`

graph{y=-2x^2+5x-1 [-10, 10, -5, 5]}

Answer 2

`x = -0.22 " and " x = 2.28` for `y = 0`

Explanation:

When we say "solution" instead of the entire curve, we usually mean the place(s) were the function is zero. That is, the "roots" of the equation are at:

`-2x^2 + 5x - 1 = 0`

Now you can solve this be number of ways - factoring, quadratic formula, and graphing.

This one may be solved most quickly with the quadratic formula.

`x =( −b ± sqrt(b^2−4ac))/(2a)`
in this case, `a = -2, b = 5, c = -1`

`x =( −5 ± sqrt(5^2−4(-2)(-1)))/(2(-2))`

`x = (−5 ± sqrt(17))/(−4)`

`x = 5/4 ± sqrt(17)/4` ; `x = 5/4 ± 1.03`

`x = -0.22 " and " x = 2.28`