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kirk.freedman
Group Title
Use the quadratic formula to solve the equation.
x^2 – 7x – 6 = 0
Show work please.
 2 years ago
 2 years ago
kirk.freedman Group Title
Use the quadratic formula to solve the equation. x^2 – 7x – 6 = 0 Show work please.
 2 years ago
 2 years ago

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winterfez Group TitleBest ResponseYou've already chosen the best response.0
x^27x6=0?
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
\[x = \frac{ b \pm \sqrt{b ^{2} 4ac} }{ 2a}\]
 2 years ago

kirk.freedman Group TitleBest ResponseYou've already chosen the best response.0
yes it is an exponent
 2 years ago

lilg132 Group TitleBest ResponseYou've already chosen the best response.0
x2 = a 7x = b 6 = c x2 = 1 7x = 7 6 = 6
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
\[x = \frac{ b \pm \sqrt{b ^{2}4ac} }{ 2a }\]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.
 2 years ago

kirk.freedman Group TitleBest ResponseYou've already chosen the best response.0
Hmm, ok thanks, let me work this out and I will reply to what I got to check if it is right.
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

winterfez Group TitleBest ResponseYou've already chosen the best response.0
dw:1354234621716:dw
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
Are you able to identify the variables a, b, and c and make the substitutions?
 2 years ago

kirk.freedman Group TitleBest ResponseYou've already chosen the best response.0
Yes, I got to the step that winterfezz last showed, but I don't know what to do now
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
Start with the expression within the radical. Start with (7)^2. That is the same as 7^2.
 2 years ago

winterfez Group TitleBest ResponseYou've already chosen the best response.0
dw:1354234898685:dw you get 2 answers
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
And when you have a negative number times a positive number, the product will be negative.
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
So, can you determine (7)^2 ?
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
(7)^2 is the same as (7)(7) and that is the same as (7)(7)
 2 years ago

kirk.freedman Group TitleBest ResponseYou've already chosen the best response.0
Yes, tcarroll, I know all that basic math stuff xD I am at the part where I have dw:1354235055422:dw
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
That's good!. Because that is your final answer! Well done!
 2 years ago

kirk.freedman Group TitleBest ResponseYou've already chosen the best response.0
So I do not simplify it further?
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
That's as far as you can take it.
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
It is customary to leave it with the + sign
 2 years ago

kirk.freedman Group TitleBest ResponseYou've already chosen the best response.0
Ok, so no decimal needed, I just leave it as this?
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
That's good just like that.
 2 years ago

kirk.freedman Group TitleBest ResponseYou've already chosen the best response.0
Thank both of you very much!:)
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ 7\pm \sqrt{73} }{ 2 }\]Like that
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.1
You're very welcome!
 2 years ago
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