Please Helppp !!!! Solve 5x2 = -30x - 65. x = -3 ± 2i x = -3 ± 4i x = -6 ± 2i x = -6 ± 4i
Do you know the Quadratic formulae?
yes, i do know the quadratic formula
\[x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a}\]
Rearrage your eqn in quadratic form \[ax^2 + bx + c = 0\]subst a, b, c and solve for x
hold on , let me try
i'm kind of confused
Which part?
Putting all together : \[\text{Putting all together: } x = \frac{ -30\pm \sqrt{30^2 - 4*5*65}}{ 2*5}\]
\[ax ^{2}+bx+c+0 \] When i substitute them i get \[5x ^{2}+30x-+65=0\] ?
When you move the expression on the right to the left, the -ve signs turn +ve. You remember?
yes,
\[\text{So you get : } 5x ^{2}+30x +65=0\]
i also remember that when solving the quadratic formula there was something that i have to do first, and i can't remember if it's the bottom half of the equation that i start with, or the part that says 30^2 - 4*5*65
\[\text{Simplify this further: } x = \frac{ -30\pm \sqrt{30^2 - 4*5*65}}{ 2*5} \rightarrow x = \frac{ -30\pm \sqrt{900 - 1300}}{10}\]
Does that make sense now?
\[x = \frac{ -30\pm \sqrt{900 - 1300}}{10} = \frac{ -30 }{ 10 } \pm \frac{ \sqrt{-400} }{ 10 }\]
and next you do 900 - 1300 and get \[x= -30\pm \sqrt{-400} / 10\]
or do i make the 400 positive ?
Nope, that's where the imaginary number 'i' comes in, since we do not the negative square root of a number...
ok
You have it rather like this\[x = -3 \pm \frac{ \sqrt{400i} }{ 10 } \rightarrow x = -3 \pm \frac{ 20i }{ 10 } \]
\[x = -3 \pm 2i\]
wow that is confusing, haha
It's very important you understand every bit of the process that got us here. So, what confusion can we clear up now?
the imaginary number is whats confusing me
Yeah, it's just a rule. You know that now, it won't confuse you any more :)
ohh i see, the 0's cancel out
Yeah
And you notice that we took the sqrt of 400 too
wait, what happened with the sort of 400 ? it canceled out too ?
You just keep the i out of it and take the square root of its coefficient.
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