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myininaya Group TitleBest ResponseYou've already chosen the best response.2
add 6x to both sides
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
\[2x^2+6x5=0\]
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Now use the quadratic formula! :)
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
a=2,b=6,c=5
 2 years ago

qudrex Group TitleBest ResponseYou've already chosen the best response.0
whats the quadratic formula?
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
\[x=\frac{b \pm \sqrt{b^24ac}}{2a}\]
 2 years ago

qudrex Group TitleBest ResponseYou've already chosen the best response.0
i dont know how to use that. D:
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Do you know how to complete the square?
 2 years ago

qudrex Group TitleBest ResponseYou've already chosen the best response.0
no. :/ im horrible at math.
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
hmmm... so what method do you want to use?
 2 years ago

qudrex Group TitleBest ResponseYou've already chosen the best response.0
i dont know any methods to solve that.
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
@Qudrex I'll help you.
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
A standard quadratic equation is given as \[ax^2+bx+c=0\] its roots or its solution is given by the following formula \[x=\frac{b \pm \sqrt{b^24ac}}{2a}\]
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
This formula gives us two values of x which satisfies the quadratic equation
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
Did you understand till now?
 2 years ago

qudrex Group TitleBest ResponseYou've already chosen the best response.0
formulas confuse me.
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
what's confusing you?
 2 years ago

qudrex Group TitleBest ResponseYou've already chosen the best response.0
the formula i dont know how to use it.
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
It's very easy, I'll show you your question is 5+2x^2=6x The first step is to bring all the terms to one side \[5+2x^2=6x\] Let's subtract both sides by 6x we get \[5+2x^26x=6x6x\] we'll get \[5+2x^26x=0\] Now let's write this in descending powers of x \[2x^26x5\] Did you understand till now?
 2 years ago

qudrex Group TitleBest ResponseYou've already chosen the best response.0
yes. i think i understand that.
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
Great now let's compare this with standard equation \[ax^2+bx+c=0\] \[2x^26x+5=0\] we have \[a=2,\ b=6\ and\ c=5 \] Did you get this?
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
Now its solution is given by \[x=\frac{b\pm \sqrt{b^24ac}}{2a}\] Let's substitute a=2, b=6 and c=5 \[x=\frac{(6)\pm \sqrt{(6)^24\times (2)\times (5)}}{2\times (2)}\] We get now \[x=\frac{6\pm \sqrt{36(40)}}{4}\] \[x=\frac{6\pm \sqrt{36+40}}{4}\] \[x=\frac{6\pm \sqrt{76}}{4}\] Did you understand this?
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
We've almost found the solution, we could simplify it more We have \[x=\frac{6\pm \sqrt{76}}{4}\] \[\sqrt {76}= \sqrt{4\times 19}=2\sqrt {19}\] so \[x=\frac{6\pm 2\sqrt{19}}{4}\] divide the whole equation by 2 we get \[x=\frac{3\pm \sqrt {19}}{2}\] so the two solutions\roots are \[x=\frac{3+ \sqrt {19}}{2}, \frac{3 \sqrt {19}}{2}\]\] Did you understand?
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.2
So whenever you've to find roots of quadratic equation, use this formula:D
 2 years ago
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