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

N

nae

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Jun 18, 2006
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I need help with these problems:

1. x(7x-4)=0

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

I am lost.
 
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.
 
nae said:
2. x^2-3x+9x=27
Click to expand...
That equation is same as: x^2 + 6x - 27 = 0
Do you understand why?
If not, not much can be done for you here.
 
stapel said:
The first one is already in factored form. .
Click to expand...
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?
 
nae said:
2. x^2-3x+9x=27

I am lost.
Click to expand...

No need to be. Just listen up :wink:

\(\displaystyle \L\bold x^2-3x+9x=27\)
Can be simplified as: \(\displaystyle \L\bold x^2+6x=27\)

In order to solve this by using the Quadratic Formula you need to have it in this form:
\(\displaystyle \L\bold ax^2+bx+c=0\) In which \(\displaystyle \L\bold a\) is unequal to zero.

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

Now use the quadratic forumula: \(\displaystyle \L\bold
x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}\)

You will get two answers.
 
jonboy said:
stapel said:
The first one is already in factored form. .
Click to expand...
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?
Click to expand...
No need for quadratic, jonboy:
x(7x - 4) = 0
x = 0 or 7x - 4 = 0
x = 0 or x = 4/7
 
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
 
mcrae said:
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
Click to expand...

When did I do that?

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

jonboy said:
Since you can put \(\displaystyle \bold p^2+6p+4=0\) in the form \(\displaystyle ax^2 + bx + c = 0\) in which a is not=0, yes it can be factored by using the Quadratic Formula.
Click to expand...

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

jonboy said:
As long as \(\displaystyle a\) is not equal to zero and can be expressed using \(\displaystyle ax^2+bx+c=0\) then it can be solved by using the Quadratic Formula.
Click to expand...
 
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.
 
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