Question

Approximate the real zeros to the nearest tenth -x^2+x+4

Answer 1

The zeros of the given expression at x = -1.6 and x = 2.6

Explanation:

Given: -x^2 + x + 4

To find the real zeros set the expression equal to zero and solve for x.

-x^2 + x + 4 = 0

This cannot be factored.

Let's use the quadratic formula and find the solutions.

Formula x =
`(-b +/-√(b^2 - 4ac) )/(2a)`

Here a = -1, b = 1 and c = 4 from the given expression.

Now let's plug in these values in the above formula, we get

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

x =
`(-1 +/- √(1 + 16) )/(-12)`

x = (-1 +/-
`√(17)`)/-2 [√17 = 4.1]

x = (-1 + 4.1)/ -2 and x = (-1 -4.1)/-2

x = 3.1/-2 and x = -5.1/-2

x = -1.55 and x = 2.55

When we round off to the nearest tenths place, we get

x = -1.6 and x = 2.6

Therefore, the zeros of the given expression at x = -1.6 and x = 2.6.

Hope this will helpful.

Thank you.