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Quadratic equation
- Thread starter davehogan
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I'm stuck with a factoring problem, 4x^2-11x+6=0. I get to 2x(2x-4)-3(x-2) but cannot get to the answer: x=2 or 3/4. Help!
For a quick review - go to:
http://www.purplemath.com/modules/solvquad.htm
First of all, factoring is not the easiest way to solve most quadratic equations. If you "see" a factoring, then it is quickest, but if you don't "see" one, then use one of the other methods for solving it UNLESS OF COURSE THE PROBLEM REQUIRES A SOLUTION BY FACTORING.I'm stuck with a factoring problem, 4x^2-11x+6=0. I get to 2x(2x-4)-3(x-2) but cannot get to the answer: x=2 or 3/4. Help!
Second, I am not sure how you got to where you got, but you are good so far. Do you see that (2x - 4) can be factored further? Try that, and
I think you will be able to complete the factoring
I have no idea what you are doing because you are not showing your work.Yes, that gives me x=2. What about the other side which apparently gives x=3/4?
Do you understand the zero property concept behind solving a quadratic by factoring?
Let's take an example
\(\displaystyle x^2 - x - 42 = 0.\)
I can factor that equation so that it becomes a product of two linear terms.
\(\displaystyle x^2 - x - 42 = 0 \implies (x - 7)(x + 6) = 0.\) Does that make sense to you?
If a * b = 0, then either a = 0 or b = 0 or a = b = 0. This is the zero product property. So let's apply it to the equation in the line above.
\(\displaystyle (x - 7)(x + 6) = 0 \implies x - 7 = 0\ or\ x + 6 = 0 \implies x = 7\ or\ x = - 6.\)
Now as far as I can see you have not reduced your quadratic to a simple product of two linear terms. Try doing that and then applying the zero product property.
By the way, this zero product property is a way to find the zeroes of polynomials with a higher degree than 2 such as cubics, quartics, qunitics, etc. So it is a useful technique to learn, but you never have to use it with quadratics.
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I'm stuck with a factoring problem, 4x^2-11x+6=0. I get to 2x(2x-4)-3(x-2) but cannot get to the answer: x=2 or 3/4. Help!
If (x-2) is one factor, you can divide each term in the expression by (x-2), and what is left is the other factor.Yes, that gives me x=2. What about the other side which apparently gives x=3/4?
I'm stuck with a factoring problem, 4x^2-11x+6=0. I get to 2x(2x-4)-3(x-2) = 0
but cannot get to the answer: x=2 or 3/4. Help!
davehogan, please answer this if you would. Did you have the quadratic equation expressed as "4x^2 - 8x - 3x + 6 = 0"
before you got it to 2x(2x - 4) - 3(x - 2) = 0?
You are not going to get x = 3/4 along this route until you get back on track with the correct factor by grouping method.
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The original equation was 4x^2-11x+6. I multiplied the coefficient of x^2 by the variable making 24. Thereafter as follows:
4x^2-11x+6=0
4x^2-8x-3x +6=0
2x(2x-4)-3(x-2)=0
(2x-4)????????
2x-4 goes on to make x=2. But I can't see how to go on to get 3/4.
4x^2-11x+6=0
4x^2-8x-3x +6=0
2x(2x-4)-3(x-2)=0
(2x-4)????????
2x-4 goes on to make x=2. But I can't see how to go on to get 3/4.
The original equation was 4x^2-11x+6. I multiplied the coefficient of x^2 by the variable making 24. Thereafter as follows:
4x^2-11x+6=0
4x^2-8x-3x +6=0
2x(2x-4)-3(x-2)=0
(2x-4)????????
2x-4 goes on to make x=2. But I can't see how to go on to get 3/4.
Since you insist to factor this way,
\(\displaystyle
2x(2x-4)-3(x-2)= 2x(2x-4)-\frac{3}{2}(2x-4)\)
HallsofIvy
Elite Member
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- Jan 27, 2012
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And then \(\displaystyle (2x- \frac{3}{2})(2x- 4)\)Since you insist to factor this way,
\(\displaystyle
2x(2x-4)-3(x-2)= 2x(2x-4)-\frac{3}{2}(2x-4)\)
But your basic problem, DaveHogan, is not noticing that 2x- 4= 2(x- 2). A better factoring is
4x(x- 2)- 3(x- 2)= (4x- 3)(x- 2).
I explained all this before. Did you even read my previous posts?The original equation was 4x^2-11x+6. I multiplied the coefficient of x^2 by the variable making 24. Thereafter as follows:
4x^2-11x+6=0
4x^2-8x-3x +6=0
2x(2x-4)-3(x-2)=0
(2x-4)????????
2x-4 goes on to make x=2. But I can't see how to go on to get 3/4.
\(\displaystyle 2x(2x - 4) - 3(x - 2) = 2x(2)(x - 2) - 3 (x - 2) = 4x(x - 2) - 3(x - 2) = what?\)
Where did the minus 3/2 come from? To solve a quadratic by factoring you need to reduce it to a single product of two linear terms in x.But surely -3/2(2x-4) gives us x=2 again. I'm supposed to get 3/4.