x² - 3x = 10
Move the 10 to the left hand size by subtracting 10 from both:
x² - 3x - 10 = 0
Now think of factors of -10 that add to be -3
-5 x 2 = -10
-5 + 2 = -3
So the answer is:
(x - 5)(x + 2) = 0
This means either (x-5) is zero, or (x+2) is zero. Therefore:
x = 5 or x = -2
2007-01-17 08:54:52
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answer #1
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answered by Puzzling 7
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Use the person-friendly actuality that 10^log(x) = x use the two facets because of the fact the exponent of 10 (i'm assuming the log is base 10) sqrt(x^2 - 3x) = 10^(a million/2) = sqrt(10) sq. the two facets x^2 - 3x = 10 x^2 - 3x - 10 = 0 element (x-5)(x+2) = 0 x=5 or x=-2
2016-10-31 09:26:20
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answer #2
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answered by nocera 4
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(1) What are the factors of 10?
1, 10, 2, 5.
YOu need to arive at a middle number of "-3x" when you multiply the two factors, so you're probably not going to use 10 and 1, instead 2 & 5.
So It's
(x(+ or -) 5) and (x (+ or -)2)
Which one, when multiplied together, will get you -3?
-5 and +2
So you have (x-5) and (x+2)
Set them so they equal to zero---
x = 5, -2
2007-01-17 08:56:11
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answer #3
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answered by Perdendosi 7
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1. get the ten on the left x^2-3x-10=0
2. factor (X+5)(x-2)=0
3. resolve x=-5 x=2
2007-01-17 08:55:40
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answer #4
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answered by jake78745 5
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x-3x=10
x(1-3)=10
x(-2)=10
-2x=10
-x=10/2
-x=5
x=-5
substituting x as -5,we get
15=10(which is not possible)
this shows that the equation is wrong
kindly excuse if i am wrong..i am a commerce student
2007-01-17 08:58:33
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answer #5
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answered by sam 4
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x^2-3x=10
First subtract 10 from both sides
x^2-3x-10 = 0
Now try to unfoil
(x-5)(x+2) = 0
Now set each to = 0
x-5 = 0 or x+2 = 0
x=5 or x = -2
2007-01-17 08:55:36
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answer #6
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answered by Anonymous
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x(x-3)=10
multiple eveything in the parenthesis by 3
x*x=x ^2 then multiply -3 times x and you get -3x and you leave the 10 the way it is.....
Combine like terms which would equal to -2x
Your problem should look like this -2x=10
Now divide both sides by 2 and you'll get -5
2007-01-17 08:53:32
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answer #7
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answered by ~Zaiyonna's Mommy~ 3
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x^2-3x=10
x^2-3x-10=0
find two numbers a+b=-3 a*b=-10
a&b should be -5&2
(x-5)(x+2)=0
2007-01-17 08:53:14
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answer #8
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answered by Anonymous
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x^2 - 3x -10 = 0
(x-5)(x+2) = 0
x-5 = 0 or x+2 = 0
x = 5 or -2
2007-01-17 08:54:14
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answer #9
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answered by mortgagelns 3
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x^2-3x=10
x^2-3x-10=0
(x-(3/2))^2-(49/4)=0
(x-(3/2))^2-(49/4)=0
(x-(3/2)-(7/2))(x-(3/2)+(7/2))
therefore (x-(3/2)-(7/2)) or (x-(3/2)+(7/2) is equal to zero
u do the rest
2007-01-17 09:04:31
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answer #10
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answered by Anonymous
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