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. 3x^2 - 10 x + 15 = 0

2007-07-04 04:54:46 · 9 answers · asked by Anonymous in Science & Mathematics Mathematics

so whats the complex roots

2007-07-04 05:07:06 · update #1

thx all 4 hlp me ...

2007-07-04 05:15:59 · update #2

9 answers

3x^2 - 10 x + 15 = 0

Use the quadratic formula:
[-b +/- sqrt(b^2 - 4ac)] / 2a
= [10 +/- sqrt(100 - 4*3*15)] / 2*3
= [10 +/- sqrt(100 - 180)] / 6
= [10 +/- sqrt(-80)] / 6

So there is no real solution since you are taking the square root of a negative number, but you can find the complex roots:

[10 +/- sqrt(-80)] / 6
= [10 +/- 4*sqrt(-5)] / 6
= [5 +/- 2*sqrt(-5)] / 3
= [5 +/- 2i*sqrt(5)] / 3

2007-07-04 05:04:24 · answer #1 · answered by whitesox09 7 · 0 0

Hey there!

I thought I had answered this question. Guess you have deleted it. I'll solve the equation again, by completing the square.

3x^2-10x+15=0 -->
3x^2-10x=-15 -->
x^2-(10/3)x=-5 -->
x^2-(10/3)x+(10/6)^2=-5+(10/6)^2 -->
x^2-(10/3)x+(5/3)^2=-5+(5/3)^2 -->
(x-5/3)^2=(-45/9)+(25/9) -->
(x-5/3)^2=-20/9 -->
x-5/3=±sqrt(-20/9) -->
x-5/3=±sqrt(-20)/sqrt(9) -->
x-5/3=±2isqrt(5)/3 -->
x=5/3±2isqrt(5)/3 -->
x=(5±2isqrt(5))/3

So the answer is x=(5±2isqrt(5))/3 or x is approximately equal to 1.67±1.49.

Hope it helps!

2007-07-04 12:18:00 · answer #2 · answered by ? 6 · 0 0

This is a simple quadratic polynomial in the form:

ax^2 + bx + c

where a=0.3, b=-10, c=15

To find the roots of this equation, use the quadratic equation:

[-b +- sqrt(b^2 - 4ac)] / 2a

[10 +- sqrt(100 - (4)(0.3)(15))] / (0.3)(2)

[10 +- sqrt(100-18)] / 0.6

[10 +- sqrt(82)] / 0.6

So, there are two roots of the equation:

[10 + sqrt(82)] / 0.6 and [10 - sqrt(82)] / 0.6

The square root of 82 is about 9.05538514, so the two roots are

31.7589752 and 1.5743581

Thus, that equation is true when x is either 31.7589752 or 1.5743581.

2007-07-04 12:16:03 · answer #3 · answered by Anonymous · 0 1

Standard form: ax^2 + bx + c = 0
3x^2 - 10x + 15 = 0
where
a = 3, b = -10, c = 15

By the Quadratic formula,

x = [ -(-10) + or- sqrt( (-10)^2 - 4(3)(15) ] / [ 2(3) ]

x = [ 10 +or- sqrt(-80) ] / 6

note: sqrt(-80) = sqrt(16)sqrt(-1)sqrt(5) = 4i*srqt(5)

x = [ 10 +or- 4i*sqrt(5) ] / 6

x = 2[ 5 +or- 2i*sqrt(5) ] / (2*3)

x = [ 5 + 2i*sqrt(5) ] / 3 or x = [ 5 - 2i*sqrt(5) ] / 3

Answer: x = { [ 5 + 2i*sqrt(5) ] / 3, x = [ 5 - 2i*sqrt(5) ] / 3 }

2007-07-04 12:12:49 · answer #4 · answered by mathjoe 3 · 0 0

3x² - 10x + 15 = 0
x = [ 10 ± √ (- 80) ] / 6
x = [ 10 ± 4√5 i ] / 6
x = [ 5 ± 2√5 i ] / 3

2007-07-04 14:32:49 · answer #5 · answered by Como 7 · 0 0

3x^2 - 10 x + 15 = 0
Use the quadratic formula
x= [-(-10) +/- sqrt((-10)^2-4(3)(15))]/(2*3)
x = [10 +/- sqrt(100 - 180)]/6
x = [10 +/- sqrt(-80)]/6
x = [10 +/- 4sqrt(5)i]/6
x = 5/3 +/- (2i/3)*sqrt(5)

2007-07-04 12:06:17 · answer #6 · answered by ironduke8159 7 · 0 0

Kyle and Ironduke appear correct. If you just copy their solution into your homework you'll get the answer right, but you won't learn it so you'll miss the same question on the test (when it counts). So follow what they did and re-work it by hand yourself.

2007-07-04 12:17:11 · answer #7 · answered by mike 3 · 2 0

3x^2-10x+15=0
-15 -15
3x^2-10x=-15
I don't know what the ^ symbol means so I can't do the rest.

2007-07-04 12:02:23 · answer #8 · answered by ...... 3 · 0 2

no real root babe.....dont try to fool around i know even u that well dont u...?!1

2007-07-04 12:03:35 · answer #9 · answered by Ξlectronegative™дtif® 2 · 0 2

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