Question

the solutions of x^2+bx+c=0 are each 5 more than the solution of x^2+7x+3=0.
what is the value of b and c

Answer 1

That is very interesting. Is there a question hidden here somewhere?

Answer 2

I would just solve the x^2 + 7x + 3 and take the two roots and add 5 to them. With that sum I would take the product of (x and these values) being careful with signs. I am gonna try that.

Answer 3

\[x ^{2}+7x+3=0\] Using quadratic formula:\[-7\pm \sqrt{7^{2}-4(1)(3)}\over2\]\[-7\sqrt{49-12}\over2\]\[-7\pm \sqrt{37}\over2\] Now calculator time

Answer 4

this really helped

Answer 5

x=-0.4586
x=-6.5414

Answer 6

add 5 to each of these roots
Now use these factors (x+5.4586)(x+11.5414) should result in a new equation giving you the requested b and c values. Your professor must have a mean streak in him. lol

Answer 7

ERROR ALERT. I messed up with the signs when I added 5.

Answer 8

Since they were negative when adding 5 I will have 5-0.4586 =4.5414
and 5-6.5414=-1.5414, so the product would be (x-4.5414)(x+1.5414)

Answer 9

So the new equation would be:\[x ^{2}-3x -7\]

Answer 10

b=-3, and c=-7

Answer 11

thank

Answer 12

Do you have and option for answers and is that one?