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- Thread starter mb9595
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D
Deleted member 4993
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mb9595 said:solve for r
the square root of (3r +1)=2r=6 <<< This could not be correct - fix it
mb9595 said:its radical[3r+ 1]=2r+6
[3r+1] are the numbers under the radical
Heres my work
3r+1=(2r+6)(2r+6)
3r+1=(4r^2+24r+36)
0=(4r^2=24r=36)-3r-1
I don't know what to do with it
Sorry I mistype it orginally
Squaring both sides is the "right" thing to do.
BUT.....your results need some correction, too.
THIS.....
0=(4r^2=24r=36)-3r-1
has three equals signs. That can't be right.
D
Deleted member 4993
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\(\displaystyle 4r^2 \ + \ 21r \ + 35 \ = \ 0\)
This is a quadratic equation. What methods have you been taught to solve these?
This is a quadratic equation. What methods have you been taught to solve these?
I think the answer might be no real solutions. We never had a problem like this before. I know what the quadratic formula is. I tried this problem again and my answer had an imaginary number in it so it much be no real solution. Thank-you for your time trying to help me.
D
Deleted member 4993
Guest
mb9595 said:I think the answer might be no real solutions. <<<< Correct
We never had a problem like this before. I know what the quadratic formula is. I tried this problem again and my answer had an imaginary number in it so it much be no real solution. Thank-you for your time trying to help me.