I can't dp this for the life of me
please help
b^2+13b^2+36=0
2007-10-21 04:15:50 · 4 answers · asked by sumguy 1 in Science & Mathematics ➔ Mathematics
b^2+13b^2+36=14b^2+36=0 So b*b= -36/14=-18/7 The only way you can have a number multiplied by itself be negative is if it is an Imaginary number. b=i sqrt[18/7]=i 3*sqrt[2/7]
If you mistyped the problem and it actually is: b^2+13b+36=0 Then the solution is:
Notice that 36 is the product of two numbers. 1x36 and 2x18 and 4x9 and 6x6 all give 36. The 13 in the equation is the sum of the same numbers that when multiplied give 36. NOTICE:4+9=13 so 4 and 9 are your roots! Check:
(b+4)(b+9)=b^2+9b+4b+36=
b^2+13b+36 so it works. Really helps if you know your multiplication tables. So the roots are REAL, not imaginary and are x=-4 and x=-9
2007-10-21 04:36:28 · answer #1 · answered by oldschool 7 · 0⤊ 0⤋
So the equation is 14 b^2 +36 = 0 ?
2007-10-21 04:20:55 · answer #2 · answered by e2theitheta 2 · 0⤊ 0⤋
I believe you copied the problem incoeeectly
b^2+13b+36=0 factors easily.
2007-10-21 04:35:40 · answer #3 · answered by Anonymous · 0⤊ 0⤋
suppose b^2+13b+36 =0
b=((-13+-sqrt(169-144))/2 so b=-9 and b= -4
2007-10-21 04:24:11 · answer #4 · answered by santmann2002 7 · 0⤊ 0⤋