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Check this out

http://www.wolframalpha.com/input/?i=3x%5E2-12x%2B19

im not sure if this is what you want, any way that you can solve it in a TI calculater like this: Solve(3x^2-12x+19,x)

that will give you the answer for x

Posted on Sep 19, 2013

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Posted on Jan 02, 2017

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Let x be the smallest number.

Let x + 2 be the other number (consecutive even integer)

Now to translate the rest;)

three times the smaller 3(x)

19 more -19

sum of the two integers - (x) +( x+2)

Pulling it together,

3x -19 = x + x +2

collect like terms

3x - 19 = 2x + 2

Put all the constants on one side by adding 19 to both sides.

3x - 19 +19= 2x + 2 + 19

3x = 2x +21

Subtract 2x from both sides to have all the x's on one side.

3x - 2x = 2x +21 - 2x

x = 21

The other number is 21 + 2, or 23

Check:three times the smaller = 3 x 21 = 63

sum of the two integers = 21 + 23 or 44

is 63 at least 19 more than 44

Let x + 2 be the other number (consecutive even integer)

Now to translate the rest;)

three times the smaller 3(x)

19 more -19

sum of the two integers - (x) +( x+2)

Pulling it together,

3x -19 = x + x +2

collect like terms

3x - 19 = 2x + 2

Put all the constants on one side by adding 19 to both sides.

3x - 19 +19= 2x + 2 + 19

3x = 2x +21

Subtract 2x from both sides to have all the x's on one side.

3x - 2x = 2x +21 - 2x

x = 21

The other number is 21 + 2, or 23

Check:three times the smaller = 3 x 21 = 63

sum of the two integers = 21 + 23 or 44

is 63 at least 19 more than 44

Feb 19, 2015 | Office Equipment & Supplies

It seems to me that you are trying to solve the quadratic equation

aX^2+bX+c=10 with a=-3, b=3, c=15 or**-3X^2+3X+15=0**.

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation**-3X^2+3X+15=0**..

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

**Solution1 =X_1=(-3-SQRT(189))/(-2*3)=(1+SQRT(21))/2**

**Solution2 =X_2=(-3+SQRT(189))/(-2*3)=(1-SQRT(21))/2** or about -1.791287847

Here SQRT stands for square root.

aX^2+bX+c=10 with a=-3, b=3, c=15 or

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

Here SQRT stands for square root.

Aug 17, 2014 | Computers & Internet

It seems to me that you are trying to solve the quadratic equation

aX^2+bX+c=10 with a=-3, b=3, c=15 or**-3X^2+3X+15=0**.

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation**-3X^2+3X+15=0**..

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

**Solution1 =X_1=(-3-SQRT(189))/(-2*3)=(1+SQRT(21))/2**

**Solution2 =X_2=(-3+SQRT(189))/(-2*3)=(1-SQRT(21))/2** or about -1.791287847

Here SQRT stands for square root.

aX^2+bX+c=10 with a=-3, b=3, c=15 or

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

Here SQRT stands for square root.

Aug 17, 2014 | SoftMath Algebrator - Algebra Homework...

In Algebra

Likewise when you see

Special Binomial Products So when you multiply binomials you get ... Binomial Products

And we are going to look at

1. Multiplying a Binomial by Itself What happens when you square a binomial (in other words, multiply it by itself) .. ?

(a+b)2 = (a+b)(a+b) = ... ?

The result:

(a+b)2 = a2 + 2ab + b2

You can easily see why it works, in this diagram:

2. Subtract Times Subtract
And what happens if you square a binomial with a **minus** inside?

(a-b)2 = (a-b)(a-b) = ... ?

The result:

(a-b)2 = a2 - 2ab + b2

3. Add Times Subtract
And then there is one more special case... what if you multiply (a+b) by (a-b) ?

(a+b)(a-b) = ... ?

The result:

(a+b)(a-b) = a2 - b2

That was interesting! It ended up very simple.

And it is called the "**difference of two squares**" (the two squares are **a2** and **b2**).

This illustration may help you see why it works:

a2 - b2 is equal to (a+b)(a-b)
Note: it does not matter if (a-b) comes first:

(a-b)(a+b) = a2 - b2

The Three Cases
Here are the three results we just got:

(a+b)2
= a2 + 2ab + b2
} (the "perfect square trinomials")
(a-b)2
= a2 - 2ab + b2
(a+b)(a-b)
= a2 - b2
(the "difference of squares")
Remember those patterns, they will save you time and help you solve many algebra puzzles.

Using Them
So far we have just used "a" and "b", but they could be anything.

Example: (y+1)2

We can use the (a+b)2 case where "a" is y, and "b" is 1:

(y+1)2 = (y)2 + 2(y)(1) + (1)2 = y2 + 2y + 1

Example: (3x-4)2

We can use the (a-b)2 case where "a" is 3x, and "b" is 4:

(3x-4)2 = (3x)2 - 2(3x)(4) + (4)2 = 9x2 - 24x + 16

Example: (4y+2)(4y-2)

We know that the result will be the difference of two squares, because:

(a+b)(a-b) = a2 - b2

so:

(4y+2)(4y-2) = (4y)2 - (2)2 = 16y2 - 4

Sometimes you can recognize the pattern of the answer:

Example: can you work out which binomials to multiply to get 4x2 - 9

Hmmm... is that the difference of two squares?

Yes! **4x2** is **(2x)2**, and **9** is **(3)2**, so we have:

4x2 - 9 = (2x)2 - (3)2

And that can be produced by the difference of squares formula:

(a+b)(a-b) = a2 - b2

Like this ("a" is 2x, and "b" is 3):

(2x+3)(2x-3) = (2x)2 - (3)2 = 4x2 - 9

So the answer is that you can multiply **(2x+3)** and **(2x-3)** to get **4x2 - 9**

Jul 26, 2011 | Computers & Internet

This equation doesn't have whole number factors so you have to solve it by completing the square or using the quadratic equation. I used the quadratic equation and got the following answers.

[3 + sqrt(209)]/10 and [3 - sqrt(209)]/10

I hope this helps you out.

[3 + sqrt(209)]/10 and [3 - sqrt(209)]/10

I hope this helps you out.

Jul 05, 2011 | SoftMath Algebrator - Algebra Homework...

I'm assumeing the problem is 2x^2 - 19x+22 = 0, solve for x.

I don't see any easy to factor this so I'll either need to complete the square or use the quadratic equation.

Using the quadratic equation, where A = 2, B = -19, and C = 22

x = [-(-19) +/- squareroot( 19^2 - 4*2*22)] / (2*2) =

(19 +/- squareroot(185))/4 =

8.150367 or 1.349632

I don't see any easy to factor this so I'll either need to complete the square or use the quadratic equation.

Using the quadratic equation, where A = 2, B = -19, and C = 22

x = [-(-19) +/- squareroot( 19^2 - 4*2*22)] / (2*2) =

(19 +/- squareroot(185))/4 =

8.150367 or 1.349632

Feb 01, 2011 | SoftMath Algebrator - Algebra Homework...

(2x - 3y)(3x + 2y) = 2x*3x + 2x*2y - 3y*3x - 3y*2y

= 6x^2 + 4xy - 9xy - 6y^2

= 6x^2 - 5xy - 6y^2 You multiply each element in the first set of brackets by each element in the second set of brackets and then consolidate like terms and arrange them in sequence of powers, first x and then y. So:

2x by 3x = 6x squared

2x by 2y = 4xy

-3y by 3x = -9xy

-3y by 2y = 6y squared

6x squared -5xy + 6y squared

= 6x^2 + 4xy - 9xy - 6y^2

= 6x^2 - 5xy - 6y^2 You multiply each element in the first set of brackets by each element in the second set of brackets and then consolidate like terms and arrange them in sequence of powers, first x and then y. So:

2x by 3x = 6x squared

2x by 2y = 4xy

-3y by 3x = -9xy

-3y by 2y = 6y squared

6x squared -5xy + 6y squared

Dec 04, 2010 | Yahoo Computers & Internet

Factor out 3X

3X(2X-5)=0

The expression is equal to zero if one of the factors on the left side equals zero.

1st solution: 3X=0, thus X=0

2nd solution 2X-5 = 0 if X=5/2

3X(2X-5)=0

The expression is equal to zero if one of the factors on the left side equals zero.

1st solution: 3X=0, thus X=0

2nd solution 2X-5 = 0 if X=5/2

Sep 10, 2010 | Office Equipment & Supplies

You can use pen and paper and write f(x)=y, then

y=(X^2+3)/(3X^2) and isolate the variable X.

The solutions are

X=+ /- 1/Square root of (Y-1/3) with the proviso that Y-1/3 strictly positive.

If you want to use the calculator

y=(X^2+3)/(3X^2) and isolate the variable X.

The solutions are

X=+ /- 1/Square root of (Y-1/3) with the proviso that Y-1/3 strictly positive.

If you want to use the calculator

- Go to the Home screen.
- Press MODE F2 and scroll to set EXACT/APPROX to Exact or AUTO
- Use keyboard to type solve( or Press F2:Algebra and select 1:solve(
- Complete the command as follows:
- solve ( 3Y*X^2-X^2-3=0, X)
- Press ENTER

Mar 23, 2010 | Texas Instruments TI-89 Calculator

I'm not sure what calculator you wanted on:Ti 89

The Ti 84/83 you would use the solver in the math key. just have it =0

The Ti 84/83 you would use the solver in the math key. just have it =0

Sep 24, 2009 | Texas Instruments Office Equipment &...

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