Solve the equation. square root of 5x+1=6 square root of x-2 +7=-4 x2+7x+4=0 5x2=-15
1) x=7 2) x=123 3) x=-6.37, -.63 4) x=9
how did u get these answers
Would like to see how the answer to the 4th one was developed. when 9 is substitued in the equation: \[5(9)^{2}=-15\] \[(9)^{2}=-15/5=-3\] \[81\neq-3\]
Yeah, my bad I read it wrong. It's actually 3i.
Gettin there...... how about \[i \sqrt{3}\]
\[5x ^{2}= -15\] \[X ^{2}= -15\div5=-3\] \[X=\sqrt{-3}\] \[X= i \sqrt{3}\]
Sorry about that, I'm new to this and I hadn't yet figured out the equation inserts.
What method did you use on number 3. Did you use quadratic formula?
Yes. First I put it in the equation and solved, simplifying tona decimal and rounding to the nearest hundredth. I also checked the zeroes on the graph.
5x^2 = -15 x^2 = -3 take the square root x = +- i sqrt(3)
Very good. lerachia is there any more questions?
blexting thanks , so their is two answers for number 4 \[i \sqrt{3}\]\[-i \sqrt{3}\]
lerachia, do you understand how quantum torch got those answers or do you care?
Yes, when you move an even power (either the numerator or denominator) there are multiple answers. I don't know how to write that with the equation thing yet though, but the rule applies, so thanks on my part as well.
good night all.
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