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kirk.freedman
Use the quadratic formula to solve the equation. x^2 – 7x – 6 = 0 Show work please.
\[x = \frac{ -b \pm \sqrt{b ^{2} -4ac} }{ 2a}\]
yes it is an exponent
x2 = a -7x = b -6 = c x2 = 1 -7x = -7 -6 = -6
\[x = \frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]where a=1, b=-7, and c =-6 This is because your equation is in the standard form of ax^2 + bx + c = 0. So, now, just make the substitutions.
Hmm, ok thanks, let me work this out and I will reply to what I got to check if it is right.
Hint: after you make your substitutions, pay special attention to the expression within the radical (the square root sign). You will see that the numbet in the radical is positive, so you will have two real solutions. But your final solution will have a radical in it.
|dw:1354234621716:dw|
Are you able to identify the variables a, b, and c and make the substitutions?
Yes, I got to the step that winterfezz last showed, but I don't know what to do now
Start with the expression within the radical. Start with (-7)^2. That is the same as 7^2.
|dw:1354234898685:dw| you get 2 answers
And when you have a negative number times a positive number, the product will be negative.
So, can you determine (-7)^2 ?
(-7)^2 is the same as (-7)(-7) and that is the same as (7)(7)
Yes, tcarroll, I know all that basic math stuff xD I am at the part where I have |dw:1354235055422:dw|
That's good!. Because that is your final answer! Well done!
So I do not simplify it further?
That's as far as you can take it.
It is customary to leave it with the +- sign
Ok, so no decimal needed, I just leave it as this?
That's good just like that.
Thank both of you very much!:)
\[\frac{ 7\pm \sqrt{73} }{ 2 }\]Like that
You're very welcome!