English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

please be detailed

2006-07-28 10:56:10 · 15 answers · asked by franciscoeleodoro 1 in Science & Mathematics Mathematics

15 answers

X ² + 2X² - 15 = 0

3 X² -15=0

3X²= 15

X² = 5
X = +√5 , X=-√5

2006-07-28 11:12:04 · answer #1 · answered by Sona 1 · 7 0

X=11

2006-07-28 12:04:58 · answer #2 · answered by Anonymous · 0 0

Well, this might be a typo. If it's just x^2 + 2x^2 - 15 = 0,
then 3x^2 = 15 ==> x^2 = 5 ==> x = +/- sqrt(5)
(plus or minus square root of 5)

But if it's a trinomial like this: x^2 + 2x - 15 = 0
then you can use the quadratic formula or complete the square. Let's use the latter method:

x^2 + 2x = 15
x^2 + 2x + 1 = 15 + 1 = 16
(x + 1)^2 = 16
x + 1 = +/- 4
x = -1 +/- 4

and there are two answers, x = -1 + 4 = 3, and x = -1 - 4 = -5

I strongly suspect this is the problem you were supposed to solve.

2006-07-28 11:23:06 · answer #3 · answered by bpiguy 7 · 0 0

Anticipating you making a typo and actually meaning x squared plus two x less 15 equals 0. The math is x^2 +2x -15=0 becomes (x+5)(x-3)=0 when factored. Making the solutions to the problem x=3 or -5.

2006-07-28 11:08:24 · answer #4 · answered by piercesk1 4 · 0 0

I'll assume that's a quadratic equation. If so, you need to use the quadratic formula which says that for an equation of the form:

a*x^2+b*x+c=0,

That there are two solutions, being:

x = -b/(2*a)+sqrt(b^2-4*a*c)/(2*a)

the other being:

x = -b/(2*a)-sqrt(b^2-4*a*c)/(2*a)

Go from there. Also be careful, as depending on how advanced in math you are, they may be giving you equations with double roots (both x's being the same) or no roots (both x's being imaginary). The way to determine this is called the determinant, and that's the b^2-4*a*c thing under the square root. If it's positive, all roots are real, if it's zero, it's a double root, and if it's negative, both solutions are imaginary.

2006-07-28 11:03:57 · answer #5 · answered by kain2396 3 · 0 0

2x^2 + 7x - 15 = 0 The equation should be factored with the intention to be solved. in case you are able to teach that 2 factors of this trinomial multiply to create 0, then which ability the two a million or the two one among them ought to be 0. commence off with (2x - ...) and (x - ...) with the aid of fact those are the only factors of two. 3&5 is going to be extra uncomplicated than a million&15, in basic terms with the aid of fact the numbers are nearer at the same time, so attempt that one first. 5*2 = 10 - 3*a million = 7. sure, This works. (2x - 3)(x + 5) = 0 Now the two (2x - 3) = 0 or (x + 5) = 0, so we resolve for the two. x = 3/2 or x = -5.

2016-10-08 10:53:55 · answer #6 · answered by ? 4 · 0 0

Assuming this is x^2 + 2x^2 - 15 = 0.

3x^2 - 15 = 0
3x^2 = 15
x^2 = 5
x = square root of 5

2006-07-28 11:05:17 · answer #7 · answered by untitled 2 · 0 0

is that second x a multiply? You should use the * for multiply when using a problem with x in it.

2X + 2 * 2 - 15 = 0
2x + 4 - 15 =0
2x - 11 = 0
2x = 11
Hopefully you can figure it out from there.

2006-07-28 11:01:21 · answer #8 · answered by Lady 5 · 0 0

Do you mean x^2 (x squared) plus 2x^2 (2x squared) - 15?

You need to show the syntax better if you want help.

2006-07-28 11:02:32 · answer #9 · answered by Stephen B 3 · 0 0

is that 2x2 like two x squared???

cuz the answer would be square root of 5

2006-07-28 10:59:18 · answer #10 · answered by da_wolverine_2007 2 · 0 0

fedest.com, questions and answers