(The ^ symbol means squared.) Use the quadratic formula to solve. Give a solution set.
4x^=9-4x
help me with my math problem?
2008-12-31 4:08 am
回答 (8)
2008-12-31 4:32 am
✔ 最佳答案
4x^2=9-4x4x^2+4x-9=0
a=4 b=4 c=-9
x=-4+/-sqrt[(4)^2-4(4)(-9)]/2(4)
x=-4+/-sqrt[16+144]/8
x=-4+/-sqrt[160]/8
x=-4+/-sqrt(16*10)/8
x=-4+/-4sq(10)/8
x=-1/2+sqrt(10)/2 or -1/2-sqrt(10)/2 answer
x=1.081 or x=-2.0811 answer
2008-12-31 8:56 pm
4x² = 9 - 4x
x² + x = 9/4
x² + 1/2x = 9/4 + (1/2)²
x² + 1/2x = 9/4 + 1/4
(x + 1/2)² = 5/2
x + 1/2 = 1.5811388
x = 1.5811388 - 0.5, x = 1.0811388
x = - 1.5811388 - 0.5, x = - 2.0811388
Answer: x = 1.0811388, - 2.0811388
Proof (x = 1.0811388):
4(1.0811388²) = 9 - 4(1.0811388)
4(1.1688611) = 9 - 4.3245552
4.67544 = 4.67544
Proof (x = - 2.0811388):
4(- 2.0811388²) = 9 - 4(- 2.0811388)
4(4.3311387) = 9 + 8.3246
17.3246 = 17.3246
x² + x = 9/4
x² + 1/2x = 9/4 + (1/2)²
x² + 1/2x = 9/4 + 1/4
(x + 1/2)² = 5/2
x + 1/2 = 1.5811388
x = 1.5811388 - 0.5, x = 1.0811388
x = - 1.5811388 - 0.5, x = - 2.0811388
Answer: x = 1.0811388, - 2.0811388
Proof (x = 1.0811388):
4(1.0811388²) = 9 - 4(1.0811388)
4(1.1688611) = 9 - 4.3245552
4.67544 = 4.67544
Proof (x = - 2.0811388):
4(- 2.0811388²) = 9 - 4(- 2.0811388)
4(4.3311387) = 9 + 8.3246
17.3246 = 17.3246
2008-12-31 4:40 pm
4x^2 = 9 - 4x
4x^2 + 4x - 9 = 0
x = [-b 屉(b^2 - 4ac)]/2a
a = 4
b = 4
c = -9
x = [-4 屉(16 + 144)]/8
x = [-4 屉160]/8
x = [-4 屉(4^2 * 10)]/8
x = [-4 ±4â10]/8
x = [-1 屉10]/2
â´ x = [-1 ±â10]/2
4x^2 + 4x - 9 = 0
x = [-b 屉(b^2 - 4ac)]/2a
a = 4
b = 4
c = -9
x = [-4 屉(16 + 144)]/8
x = [-4 屉160]/8
x = [-4 屉(4^2 * 10)]/8
x = [-4 ±4â10]/8
x = [-1 屉10]/2
â´ x = [-1 ±â10]/2
2008-12-31 12:27 pm
4x^ = 9 - 4x
4x^ + 4x = 9
4x^ + 4x -9 = 0
We will need to find the three values of this equation: a=4 b=4 c= -9
to find when 4x^ + 4x = 0 we use this formula:
(-b +/- SQRT(b^-4*a*c)) / 2*a
(-4 +/- SQRT(4^-4*4*(-9))) / 2*4
(-4 +/- SQRT(160)) / 8
x = { -2,08 and 1,08 }
4x^ + 4x = 9
4x^ + 4x -9 = 0
We will need to find the three values of this equation: a=4 b=4 c= -9
to find when 4x^ + 4x = 0 we use this formula:
(-b +/- SQRT(b^-4*a*c)) / 2*a
(-4 +/- SQRT(4^-4*4*(-9))) / 2*4
(-4 +/- SQRT(160)) / 8
x = { -2,08 and 1,08 }
2008-12-31 12:24 pm
4x^2=9-4x
4x^2+4x-9=0
x= {-b+/-rt(b^2-4ac)} / 2a
a=4, b=4, c= -9
x={-4+/-rt[(4^2)-4(4)(-9)]} / 8
x={-4+/-rt(16+144)}/ 8
x={-4+/-rt(160)} / 8
x= {-4+/-rt(16)rt(10)} / 8
x= {-4+/-4rt(10)} / 8
x=4{-1+/-rt(10)} / 8
x= [-1+/-rt(10) / 2
Solution set x={ (-1+rt10) /2, (-1-rt10) /2}
4x^2+4x-9=0
x= {-b+/-rt(b^2-4ac)} / 2a
a=4, b=4, c= -9
x={-4+/-rt[(4^2)-4(4)(-9)]} / 8
x={-4+/-rt(16+144)}/ 8
x={-4+/-rt(160)} / 8
x= {-4+/-rt(16)rt(10)} / 8
x= {-4+/-4rt(10)} / 8
x=4{-1+/-rt(10)} / 8
x= [-1+/-rt(10) / 2
Solution set x={ (-1+rt10) /2, (-1-rt10) /2}
2008-12-31 12:23 pm
Disregard this, I messed up
2008-12-31 12:22 pm
4x^=9-4x add 4x and subract 9 on both sides to get
4x^+4x-9=0 then u can use the quadritic formula so u should get
-4+or-the sqrt of 160 all over -8
4x^+4x-9=0 then u can use the quadritic formula so u should get
-4+or-the sqrt of 160 all over -8
2008-12-31 12:18 pm
First set the equation equal to 0:
4x^2 = 9 - 4x (add 4x to both sides and subtract 9 from both sides)
4x^2 + 4x - 9 = 0
For the quadratic formula:
a = 4, b = 4, c = -9
Quadratic formula: [-b +/- sqrt(b^2 - 4ac)]/2a
[-4 +/- sqrt(4^2 - 4(4)(-9)]/2(4)
[-4 +/- sqrt(16 + 144)]/8
(-4 +/- sqrt160)/8
(-4 +/- sqrt(16*10))/8
(-4 +/- 4sqrt10)/8
x = (-1 +/- sqrt10)/2 <===ANSWER
4x^2 = 9 - 4x (add 4x to both sides and subtract 9 from both sides)
4x^2 + 4x - 9 = 0
For the quadratic formula:
a = 4, b = 4, c = -9
Quadratic formula: [-b +/- sqrt(b^2 - 4ac)]/2a
[-4 +/- sqrt(4^2 - 4(4)(-9)]/2(4)
[-4 +/- sqrt(16 + 144)]/8
(-4 +/- sqrt160)/8
(-4 +/- sqrt(16*10))/8
(-4 +/- 4sqrt10)/8
x = (-1 +/- sqrt10)/2 <===ANSWER
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