i am completely stuck can someone please help me. Solve. 3x2 = –21x – 15 a.7+-3sqrt69/2 b.7+-3sqrt29/2 c.-7+-sqrt29/2 d.-7+-isqrt69/2
@Nurali
@sumi29
Rearrange your equation so everything is on one side of the equals sign, and a 0 on the other.
Having done that, see if there are any common factors to all 3 terms. If there is one, divide it out to make the rest of the problem simpler — it won't change the answer.
ok thanks
what do you get after you do that?
well i was between c and d and i think it is d which is \[-7+i \sqrt{69}/2\]
Let's work through it. What do you have for an equation after doing my suggested steps?
hold on i threw away the paper by accident when cleaning up my work space i have to rewrite it
I'll wait, no problem.
i got \[3x ^{2}+21x+15\]
=0, right? Now is there a common factor to all 3 of those terms?
yes i forgot to add the =0 to the end
common factor?
so it would be 3 as the common factor
yes. and after we take out the 3, we have?
x^2+7x+5
right. what are the values of a,b,c to put into the quadratic formula?
now im am confused
the quadratic formula says that the solutions to \[ax^2+bx+c=0,~a\ne0\] are given by \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
are you instead solving by factoring or completing the square?
I guess you can't factor here, but you could complete the square.
ok i know the second equation you put
the quadratic formula. Okay, you've got \[x^2+7x+5=0\]\[ax^2+bx+c=0\]Comparing the two, what are the needed values for \(a,b,c\) to make them equal?
zero
no...look at the two equations in my previous post. what value of \(a\) makes \(ax^2=x^2\)? what value of \(b\) makes \(bx=7x\)? what value of \(c\) makes \(c = 5\)?
by plugging in x^2 for ax^2 7x for bx and 5 for c and then i solve
i got c am i correct
@Nurali
Yes, C is the correct choice. the solutions are \[x = \frac{1}{2}(-7\pm\sqrt{29})\]
thanks for walking me thru this problem and being patient with me i am dumb when it comes to this stuff.
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