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# algebra---------------?

Select the approximate values of x that are solutions to f(x) = 0, where

f(x) = -2x2 + 5x + 7.

IS IT-

(-2.50, -3.50)

Thanks

### 3 Answers

- ?Lv 68 years agoFavorite Answer
Unfortunately, John made one slight error--in judgement, not math--after he factored the equation.

f(x) = -2x^2 + 5x + 7

f(x) = -(2x^2 - 5x - 7)

f(x) = -(x + 1)(2x - 7)

...which is really this: (-1)(x + 1)(2x - 7)

So f(x) = (-1)(x + 1)(2x - 7).

In order for f(x) to be zero, one of those factors must be zero... and since -1 is not zero, you're left with either...

(x + 1) = 0

--or--

(2x - 7) = 0

So either x = -1 or x = 7/2.

-------

Check work (something John obviously didn't do! :-)...

f(x) = -2x^2 + 5x + 7

f(-1) = -2(-1)^2 + 5(-1) + 7

f(-1) = -2(1) - 5 + 7

f(-1) = -2 - 5 + 7

f(-1) = 0

f(x) = -2x^2 + 5x + 7

f(7/2) = -2(7/2)^2 + 5(7/2) + 7

f(7/2) = -2(49/4) + 35/2 + 7

f(7/2) = -49/2 + 35/2 + 14/2

f(7/2) = 0

- navinLv 78 years ago
f(x) = - 2x^2 + 5x + 7.

=> 0 = - 2x^2 + 5x + 7

=> 2x^2 - 5x - 7 = 0

=> x^2 - 5x/2 - 7/2 = 0

=> x^2 - 2 * x * 5/4 + 25/16 = 25/16 + 7/2

=> (x - 5/4)^2 = (25 + 56)/16

=> (x - 5/4)^2 = 81/16

=> x - 5/4 = +/- 9/4

=> x = 5/4 +/- 9/4

=> x = 14/4 or - 4/4

=> x = 3.5 or - 1 Ans.

- JohnLv 78 years ago
f(x) = -2x^2 + 5x + 7 First let's factor:

f(x) = -(x+1) (2x-7) Then solve

-x+1=0

or

2x= -7

-x+1=0; -x= -1 multiply both sides by -1 to get rid of the - in front of the x

x= 1

2x=-7; x= -7/2 = -3.5

So your answers are (1, -3.5)