Question

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

Answer 1

Answer:

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

Step-by-step 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 = $\frac{-b +/-\sqrt{b^2 - 4ac} }{2a}$

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 = $\frac{-1 +/- \sqrt{1^2 - 4*-1*4} }{2*-1}$

x = $\frac{-1 +/- \sqrt{1 + 16} }{-12}$

x = (-1 +/- $\sqrt{17}$)/-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.