nae

07-06-2006, 10:13 PM

I need help with these problems:

1. x(7x-4)=0

2. x^2-3x+9x=27

I am lost.

1. x(7x-4)=0

2. x^2-3x+9x=27

I am lost.

View Full Version : solving equations: x(7x-4)=0, x^2-3x+9x=27

nae

07-06-2006, 10:13 PM

I need help with these problems:

1. x(7x-4)=0

2. x^2-3x+9x=27

I am lost.

1. x(7x-4)=0

2. x^2-3x+9x=27

I am lost.

stapel

07-06-2006, 10:19 PM

The first one is already in factored form. Has this topic not been covered in class, and that is why you're "lost", or do you know that you need to solve the two listed factors, but you have forgotten (and need links to lessons on) how to solve linear equations?

I'm going to guess that the second one has a typo. In any case, move everything over onto the left-hand side, simplify, and then apply the Quadratic Formula.

Eliz.

I'm going to guess that the second one has a typo. In any case, move everything over onto the left-hand side, simplify, and then apply the Quadratic Formula.

Eliz.

Denis

07-06-2006, 10:44 PM

2. x^2-3x+9x=27

That equation is same as: x^2 + 6x - 27 = 0

Do you understand why?

If not, not much can be done for you here.

That equation is same as: x^2 + 6x - 27 = 0

Do you understand why?

If not, not much can be done for you here.

jonboy

07-07-2006, 12:59 PM

The first one is already in factored form. .

It can be solved by using the Quadratic Formula, right? It simplies, as you know, to 7x^2-4x+0=0 which as long as a is not equal to zero then it is possible to solve, Right?

It can be solved by using the Quadratic Formula, right? It simplies, as you know, to 7x^2-4x+0=0 which as long as a is not equal to zero then it is possible to solve, Right?

jonboy

07-07-2006, 01:09 PM

2. x^2-3x+9x=27

I am lost.

No need to be. Just listen up :wink:

\L\bold x^2-3x+9x=27

Can be simplified as: \L\bold x^2+6x=27

In order to solve this by using the Quadratic Formula you need to have it in this form:

\L\bold ax^2+bx+c=0 In which \L\bold a is unequal to zero.

So you should get: \L\bold x^2+6x-27=0

Now use the quadratic forumula: \L\bold

x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}

You will get two answers.

I am lost.

No need to be. Just listen up :wink:

\L\bold x^2-3x+9x=27

Can be simplified as: \L\bold x^2+6x=27

In order to solve this by using the Quadratic Formula you need to have it in this form:

\L\bold ax^2+bx+c=0 In which \L\bold a is unequal to zero.

So you should get: \L\bold x^2+6x-27=0

Now use the quadratic forumula: \L\bold

x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}

You will get two answers.

Denis

07-07-2006, 01:38 PM

The first one is already in factored form. .

It can be solved by using the Quadratic Formula, right? It simplies, as you know, to 7x^2-4x+0=0 which as long as a is not equal to zero then it is possible to solve, Right?

No need for quadratic, jonboy:

x(7x - 4) = 0

x = 0 or 7x - 4 = 0

x = 0 or x = 4/7

It can be solved by using the Quadratic Formula, right? It simplies, as you know, to 7x^2-4x+0=0 which as long as a is not equal to zero then it is possible to solve, Right?

No need for quadratic, jonboy:

x(7x - 4) = 0

x = 0 or 7x - 4 = 0

x = 0 or x = 4/7

mcrae

07-07-2006, 03:45 PM

jonboy, you keep saying that binomials cannot be equal to 0 when using the quadratic formula, then list them as ax^2 + bx + c = 0

this was your only post where you mentioned that it is actually a that cannot be 0

just watch out for little errors like that that could completely confuse someone

this was your only post where you mentioned that it is actually a that cannot be 0

just watch out for little errors like that that could completely confuse someone

jonboy

07-07-2006, 04:05 PM

jonboy, you keep saying that binomials cannot be equal to 0 when using the quadratic formula, then list them as ax^2 + bx + c = 0

this was your only post where you mentioned that it is actually a that cannot be 0

just watch out for little errors like that that could completely confuse someone

When did I do that?

http://www.freemathhelp.com/forum/posting.php?mode=quote&p=60896

Since you can put \bold p^2+6p+4=0 in the form ax^2 + bx + c = 0 in which a is not=0, yes it can be factored by using the Quadratic Formula.

http://www.freemathhelp.com/forum/posting.php?mode=quote&p=60897

As long as a is not equal to zero and can be expressed using ax^2+bx+c=0 then it can be solved by using the Quadratic Formula.

this was your only post where you mentioned that it is actually a that cannot be 0

just watch out for little errors like that that could completely confuse someone

When did I do that?

http://www.freemathhelp.com/forum/posting.php?mode=quote&p=60896

Since you can put \bold p^2+6p+4=0 in the form ax^2 + bx + c = 0 in which a is not=0, yes it can be factored by using the Quadratic Formula.

http://www.freemathhelp.com/forum/posting.php?mode=quote&p=60897

As long as a is not equal to zero and can be expressed using ax^2+bx+c=0 then it can be solved by using the Quadratic Formula.

Denis

07-07-2006, 04:51 PM

mcrae, if a=0 in equation ax^2 + bx +c = 0, then ax^2 = 0,

and you're left with bx + c = 0 which is NOT a quadratic:

so jonboy is correct.

I wish you 2 would quit exchanging stuff that confuses everyone.

and you're left with bx + c = 0 which is NOT a quadratic:

so jonboy is correct.

I wish you 2 would quit exchanging stuff that confuses everyone.

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