1) (2x+4)^2=49
2) 2x^2-15x-7=0
thanks!
2007-09-09 12:04:58 · 4 answers · asked by boser89 1 in Science & Mathematics ➔ Mathematics
If possible i need larger and smaller values for both
2007-09-09 12:14:47 · update #1
1) 2x + 4 = 7 or 2x + 4 = -7
You will get 2 answers because these are quadratic equations.
2) use quadratic formula a=2, b= -15, c= -7
(15 +/- root 281) / 4
2007-09-09 12:15:03 · answer #1 · answered by ccw 4 · 0⤊ 0⤋
1) (2x+4)^2=49
4x² + 16x +16 = 49
4x² + 16x - 33 = 0
(2x + 11)(2x - 3) = 0
x = 1.5, x = -5.5
2) 2x^2-15x-7=0
Solve using quadratic formula
ax² + bx + c = 0
x = (-b ± â(b² - 4ac))/2a
x = (15 ± â(225 + 56))/4
x = (15 ± â(281))/4
x â 7.941, x â 0.441
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2007-09-09 19:21:31 · answer #2 · answered by Robert L 7 · 0⤊ 0⤋
expand no 1 to get 4x^2+16x+16=49, make the left side equal to zero to get 4x^2+16x-33=0 use the quadratic formula
(-b+or- sqrt b^2-4ac)/2a. x=1.5 and -5.5
Same thing for no 2. x=7.941 and -.441
2007-09-09 19:21:03 · answer #3 · answered by Rutgers101 2 · 0⤊ 0⤋
square root both sides: (2x+4)= 7
subtract: 2x=3
divide: x= 3/2
2007-09-09 19:11:54 · answer #4 · answered by kellikellikelli 2 · 0⤊ 0⤋