Quadratic equation?

Question

Can someone give me step by step instructions on how to solve this problem? It's a quadratic equation.

x^2 - 5x - 8 = 0

Answer 1

The set up for the quadratic equation is
ax^2+bx+c=0 you have: (1)x^2+(-5)x+(-8)=0
so for you a=1, b=-5, and c=-8.

There no logo here so (-+) will be plus or minus.

The quadratic equation is:
(-b (-+)(sq. root(b^2-4ac))/2a
(looks a lot prettier on paper)

Just substitute your values in now.
(-(-5) (-+)(sq. root((-5)^2-4(1)(-8)))/2(1)

(5(-+)sq. root((25+32))/2

(5(-+)sq. root((57))/2

Now you find the answer from adding and subtracting for the (-+)(plus or minus)

(5+sq. root((57))/2 = 6.275
(thats just as estimate)
(5-sq. root((57))/2 = -1.275

x = 6.275 -or- -1.275
If they want as exact answer then you use:
(5+sq. root((57))/2.

Be careful with words problems though. alot of word problems like this use an example like the length of a side of a box which cannot be negetive usually. In that case you would only use the positive answer.

Answer 2

You can use the quadratic formula:

ax^2 + bx + c = 0
... implies:
x = ( -b +/- sqrt(b^2 - 4ac)) / 2a

In your case, a=1, b=-5, c=-8. Just plug in the values and away you go:

x = ( -b +/- sqrt(b^2 - 4ac)) / 2a
x = ( -(-5) +/- sqrt((-5)^2 - 4*1*(-8)) ) / 2*1
x = ( 5 +/- sqrt(25 + 32) ) / 2
x = 5/2 +/- sqrt(57)/2

You can also solve this equation by completing the square. And you can solve some quadratic problems by looking at them and coming up with the factoring off the top of your head, though this one isn't amenable to that approach. But you didn't specify the exact means we were supposed to use.

Answer 3

It has been a while since I've dealt with quadratics (I never found a use for them) but I do remember the old saying FOIL. First, inside, outside, last. As I recall, the equation needs to be converted to (a+b)^2+(c+d)^2=0 I don't remember how to appropriately unsquare the X^2. I hope that'll jog your memory.

Answer 4

x²-5x-8=0

use the quadratic formula

x=(-b+/- (b²-4ac)^.5)/2a
x=(-(-5) +/- ((-5)²-(4*1*-8)^.5)/(2*1)
x=(5 +/- (25-(-32))^.5/2
x=(5 +/- 57^.5)/2

it reads
x equals five plus or minus the square root of 57 divided by 2

Answer 5

x² - 5x - 8 = 0

We will use the quadratic formula to solve.

ax² + bx + c = 0
x = (-b ± √(b² - 4ac))/2a

x = (5 ± √(25 + 32))/2

x = (5 ± √(57))/2

x ≈ 6.275, x ≈ -1.275
.

Answer 6

x^2-5x-8=0
Here a=1,b= -5 and c= -8
Therefore
x={-(-5)+-sqrt(25-4*1*-8)}/2*1
=(5+-sqrt(25+32)/2
=(5+-sqrt57)/2

Answer 7

x = [ 5 ± √(25 + 32) ] ∕ 2
x = [ 5 ± √57 ] / 2
This is an accurate answer and if √57 is calculated an approximate answer will be obtained.

Answer 8

x^2 - 5x - 8 = 0
a = 1, b = -5, c = -8

-b +/- sqrt (b^2 -4ac) all over 2a
-(-5) +/- sqrt [-5^2 - 4(1)(-8)] all over 2*1
5 +/- sqrt [ 25 + 32 ] all over 2
5 +/- sqrt 57 all over 2
So your answers are:
[ 5 + sqrt 57 ] / 2 ~6.27
and
[ 5 - sqrt 57 ] / 2 ~-1.27

Answer 9

quad eqn:

ax^2 + bx + c

(-b + or - sqrt(b^2 - 4ac) ) / 2a

In this case:

(5 + or - sqrt( 25 + 32 ) ) / 2

= (5 + or - 7.55) / 2

= -1.275, 6.275

Answer 10

try this website it will explain how to use the equation to solve the problem.