solve (x+3)(x-3)>0 and also solve 2x^2+5x=8?

Question

Please Show How U Did Them

Answer 1

(x + 3)(x - 3) > 0
multiply through:
x^2 + 3x - 3x - 9 > 0
x^2 - 9 > 0
x^2 > 9
x > 3 or x < -3

2x^2 + 5x = 8
2x^2 + 5x - 8 = 0
Can't factor, so you need to use quadratic formula.
a = 2
b = 5
c = -8

x = [-b +- sqrt (b^2 - 4ac)] / 2a
= [-5 +- sqrt (5^2 - 4*2*-8)] / 2(-8)
= [-5 +- sqrt (25 + 64)] / -16
= [-5 +- sqrt 89] / -16
= (5 +- sqrt 89) / 16

Answer 2

Hi,
Here's one way to do it:
1) First solve for the roots, or critical values.
x+3 = 0
x=-3
x-3 = 0
x= 3

2) Now, make a chart of signs of the factors:

x-3 - - - - - - - - - - -0 +++++++++++
x+ 3 - - - -- 0 + ++++++++++++++
--------------- -|------------|--------------
....................-3...... .....3

(Sorry, the dots on the bottom row are meaningless. They are necessary to maintain the spacing with this word processor.)
Now, multiplying the signs, you will notice that we have (-)(-) in the region between -inf and -3. That product gives +.
In the region -3 to 3 we have (-)(+). That gives a minus. Of course, the region 3 to inf. gives a plus. So, the region you want is
(-oo, -3)U(3,oo) Where I've used oo for infinity.

Prob 2:
2x² +5x = 8
2x² +5x -8 = 0
This is prime, i.e., has no integer solutions, so you must use the quadratic formula.
That is this:
x = (-b- sqrt(b²-4ac)/(2a)
This is from the notation ax² +bx +c = 0
So, in you case a=3, b=5, and c= -8.
Just plug in the numbers and crank away.
You should get something like this:
(-5+- sqrt(89))/4

Hope this helps.
FE

Answer 3

HOW to do them: For the inequality, since the product is positive, x+3 and x-3 must have the same sign (positive or negative), so find which numbers satisfy both x+3>0 & x-3>0, and then do the same for x+3<0 & x-3<0.

For the equation, get 0 on one side and then use the quadratic formula.

Answer 4

a) (x+3)(x-3)>0 <--> x<-3 or x>3 --> S=(-inf, -3) U (3, +inf)
b) 2x^2+5x=8 <--> 2x^2+5x-8=0 <-->
x=[-5±√(25+64)]/4 = [-5±√89]/4

Saludos.

Answer 5

(x + 3)(x - 3)>0

So either x + 3 > 0 or x - 3>0

x < -3 and x > +3

2x^2 + 5x = 8

2x^2 + 5x - 8 = 0

x1 = [-5 + sqrt(25 + 64)] / 4

x2 = [-5 - sqrt(25 + 64)] / 4

Answer 6

uncomplicated! you have 2x^2 + 50 = 0 we could flow the unknowns to a minimum of one side and we get X^2 = -50 /2 X^2 = -25 so X = sqrt(-25) The sqrt of a unfavorable form is a complicated form! the answer to it is (+ or - 5i) So selection 4 will artwork! desire that helps !

Answer 7

the first one...
(x+3)(x-3)>0
(x+3)(X-3)=o.
x(x+3)-3(x+3)=0
x^ + 3x -3x -9 = 0.
x^ -9 = 0
x^=9.
(square root of)9=3.
x=3.

Answer 8

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