HELP! In the middle of a college math test!?

👨🏻‍🏫 學術台數學

[匿名]

2008-12-23 3:24 pm

I'm trying to find all the real or imaginary roots of this equation! Can anyone help?!

Find all real or imaginary roots of 5x^2 + 2x + 4 = 0

更新1:

I know how to use the quadratic equation, but when I get down to (-2+-√ 76) / 10 i dont know how to finish solving it and also finding the imaginary numbers...

更新2:

Okay one more problem: same thing except with this equation... 5x^2 -x +4 = 0

回答 (10)

Engr. Ronald

2008-12-23 3:49 pm

✔ 最佳答案

5x^2+2x+4=0

we will use the quadratic formula to get its roots.
a=5 b=2 c=4
x=-b+/-sqrt[b^2-4ac]/2a
=-(2)+/-sqrt[(2)^2-4(5)(4)]/2(5)
= -2+/-sqrt[4-80]/10
= -2+/-sqrt[-76]/10
=-2+/-sqrt[4*-19]/10
=-2+/-2i[19]/10

x1=-2+2isqrt[19]/10 x2=2-2isqrt[19]/10
x1=1+isqrt[19] /5 x2=1-isqrt[19] /5 answer @(^_^)@

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[匿名]

2008-12-23 4:15 pm

5x² + 2x + 4 = 0
x² + 2/5x = - 4/5
x² + 1/5x = - 4/5 + (1/5)²
x² + 1/5x = - 20/5 + 1/25
(x + 1/5)² = - 19/25
x + 1/5 = 19i/5

x = 19i/5 - 1/5
x = - 19/5 - 1/5

Answer: x = (19i - 1)/5, - ([19i + 1]/5)

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[匿名]

2008-12-23 3:37 pm

USE QUADRATIC FORMULA...
x = (-b + -root of(b^2 - 4ac))/2a
x = (-2 + - (root of (4 -80))/10
x = (-2 + - (square of 76))/10
now 76 can be written as 4 * 19 and when u take out the square root of 4, its going to be 2....the negative side inside the square root is written as"i"...so...further
x = (-2 + - 2i root of 19)/10....now take out 2 common.....
x = (-1 + - i root of 19)/5.
so... x = (-1 + i square root of 19)/5 and x = (-1 - isquare root of 19)/5.........thats ur final answer....

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[匿名]

2008-12-23 3:47 pm

Use the Quadratic Equation:
5x² + 2x + 4 = 0
a = 5, b = 2, c = 4

x = [-b ± √(b² - 4ac)]/2a
x = [-2 ± √(2² - 4(5)(4))]/2(5)
x = [-2 ± √(4 - 80)]/10
x = [-2 ± √(-76)]/10
x = [-2 ± √(-1*76)]/10
x = [-2 ± √(-1*4*19)]/10
x = (-2 ± √-1√4√19)/10 ; i = √-1
x = (-2 ± 2i√19)/10
x = (-1 ± i√19)/5

x = (-1 + i√19)/5 = -0.2 + 0.87178i
x = (-1 - i√19)/5 = -0.2 - 0.87178i

參考: Quadratic Equation Calculator: http://www.1728.com/quadratc.htm

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An ESL Learner

2008-12-23 3:55 pm

5x^2 + 2x + 4 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 5
b = 2
c = 4

x = [-2 ±√(4 - 80)]/10
x = [-2 ±√(-76)]/10
x = [-2 ±√(2^2 * -1 * 19)]/10
x = [-2 ±2i√19]/10

∴ x = [-2 ±2i√19]/10

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[匿名]

2008-12-23 3:41 pm

[-2 +- sqrt(4-4*5*4)]/10
= [-2 +- i*sqrt(76)]/10
= [-2 +- 2i*sqrt(19)]/10

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[匿名]

2008-12-23 3:43 pm

D=2^2-4(2)(4)
=4-32
=-28 so => it's = 28 i^2
so x'=(-2-2sq7)/10 and x''=(-2+2sq7)/10 so only two imaginary roots

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cidyah

2008-12-23 3:34 pm

This equation is of form ax^2+bx+c
a = 5 b = 2 c = 4
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[-2 +/-sqrt(2^2-4(5)(4)]/(2)(5)
discriminant is b^2-4ac =-76
i^2 = -1, so √i^2 = i
No real roots: The complex roots are
x=[-2 + √(76)] / (2)(5) =[-2 + √(76)] / 10
x=[-2 - √(76)] / (2)(5) = [-2 - √(76)] / 10

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[匿名]

2008-12-23 3:34 pm

sq.rt. of (-2/5) and (-1)

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[匿名]

2008-12-23 3:33 pm

--1 plus or minus root -19 over 5

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