SOLUTION: solve using the quadratic formula and leave the answer in radical form? Could you show work as I am very confused. Thanks. Karyn y^2+6y-1=0

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: solve using the quadratic formula and leave the answer in radical form? Could you show work as I am very confused. Thanks. Karyn y^2+6y-1=0      Log On


   

Question 419764: solve using the quadratic formula and leave the answer in radical form? Could you show work as I am very confused. Thanks.
Karyn
y^2+6y-1=0
Found 2 solutions by Alan3354, Theo:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
solve using the quadratic formula and leave the answer in radical form? Could you show work as I am very confused. Thanks.
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B6x%2B-1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%286%29%5E2-4%2A1%2A-1=40.

Discriminant d=40 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-6%2B-sqrt%28+40+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%286%29%2Bsqrt%28+40+%29%29%2F2%5C1+=+0.16227766016838
x%5B2%5D+=+%28-%286%29-sqrt%28+40+%29%29%2F2%5C1+=+-6.16227766016838

Quadratic expression 1x%5E2%2B6x%2B-1 can be factored:
1x%5E2%2B6x%2B-1+=+%28x-0.16227766016838%29%2A%28x--6.16227766016838%29
Again, the answer is: 0.16227766016838, -6.16227766016838. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B6%2Ax%2B-1+%29

Sub x for y
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x = -3 ± sqrt(10)

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
equation is:

y^2 + 6y - 1 = 0

standard form of a quadratic equation is:

ax^2 + bx + c = 0

a is the coefficient of the x^2 term.
b is the coefficient of the x term.
c is the constant.

in your equation you have:

a = 1
b = 6
c = -1

the quadratic formula is:

x+=+%28-b+%2B-+sqrt%28b%5E2-4ac%29%29%2F%282a%29

you replace the a's and the b's and the c's with their respective values to get:

x+=+%28-6+%2B-+sqrt%286%5E2-4%2A1%2A-1%29%29%2F%282%2A1%29

the * is the multiplication symbol.

in the formula you see it looks like a little square.

I typed a *. The formula generator program converted it to a little square.

you would simplify this expression to be shown as follows:

x+=+%28-6+%2B-+sqrt%2840%29%29%2F2

it can also be simplified further to be shown as:

x+=+%28-3%29+%2B-+sqrt%2840%29%2F2

it can even be simplified further to be shown as:

x+=+%28-3%29+%2B-+sqrt%2810%29

sqrt%2840%29%2F2 is the same thing as sqrt%2810%29 because sqrt%2840%29 equals sqrt%282%2A2%2A10%29%2F2 which equals 2%2Asqrt%2810%29%2F2 which equals sqrt%2810%29.

since they only told you to leave the answer in radical form, then i suspect you're ok with any of the answers that still have a radical in them.

that's between you and your teacher as to how he/she wants to see the answer.

a graph of your equation looks like this:

graph%28600%2C600%2C-10%2C10%2C-20%2C20%2Cx%5E2+%2B+6x+-+1%29

you can see that the graph crosses the x-axis somewhere around x = -6.16 and x = .16 which is where the equation of x+=+%28-3%29+%2B-+sqrt%2810%29 indicates.

note that sqrt(10) is somewhere around 3.16...