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algebra
- Thread starter girlpower
- Start date
D
Deleted member 4993
Guest
If x2-5 =3x,x=?
Is your problem:
x2 - 5 = 3 * x ......................... or
x * 2 - 5 = 3 * x
Based on her other posts, girlpower seems to be studying quadratic equations right now. So I suspect that the problem is:
\(\displaystyle x^2 - 5 = 3x \implies x = what?\)
Is that correct?
What techniques does your book suggest for solving quadratics?
Which did you try?
What did you get?
\(\displaystyle x^2 - 5 = 3x \implies x = what?\)
Is that correct?
What techniques does your book suggest for solving quadratics?
Which did you try?
What did you get?
D
Deleted member 4993
Guest
yes that is correct x2-5=3x but I didn't get anything or figure it out I am confused.
It is a quadratic equation - what method/s have you been taught to solve these types of equations?
There are at least four ways to solve quadratic equations.yes that is correct x2-5=3x but I didn't get anything or figure it out I am confused.
Your book should give you at least two of them.
The first thing to do is to put the equation into standard form
\(\displaystyle x^2 - 5 = 3x \implies x^2 - 5 - 3x = 3x - 3x \implies x^2 - 3x - 5 = 0.\)
The simplest way to solve a quadratic equation in standard form is to factor the quadratic expression on the left, but there is no nice factoring in this case. If you cannot factor it easily, a way that always works is to use the quadratic formula. I'd bet a big sum of money that your book gives the quadratic formula. What is it?
Do you need help in applying the formula?
First tell me what your book says the formula isYes I need help in applying the formula.
Where in the world are these problems coming from if you have neither a book nor class materials?I don't have a book to figure out what the formula is. Can you just tell me what the formula is?
\(\displaystyle ax^2 + bx + c = 0 \implies x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\)
Note that there may be two possible distinct solutions to a quadratic equation. In some cases, one of those answers can be excluded for some reason or other. In other cases, there just are two different feasible solutions
D
Deleted member 4993
Guest
That means that if you cannot solve these - you will be placed in a class where you will be TAUGHT to tackle such problems.
So - if you cannot understand these - no harm - you'll be taught to solve these problems, as you go through school.
So - if you cannot understand these - no harm - you'll be taught to solve these problems, as you go through school.