Math help

Dobbsjr · July 4, 2016

If one of the roots of the quadratic equation 3x^2 - 9x + k = 0 is twice the other what is the value of k? I know this is easy, but I can't seem to solve it.

Tuurahk · July 4, 2016

You probably already figured it out, but in case you didn't, here's the solution I came up with:

First you need to solve 3x^2 - 9x + k = 0 to find the possible roots through the quadratic formula: x = [-b +/- sqrt(b^2 -4ac)] / 2a. So substitue things in and solve it like you would any other quadratic equation where c = k.

This will give you two possible solutions:

x = [9 + sqrt(81 - 12k)] / 6 and x = [9 - sqrt(81 - 12k)] / 6

You know that one of the roots is double the other root. So you have to create an quality like such:

[9 + sqrt(81 - 12k)] / 6 = 2 * [ [9 - sqrt(81 - 12k)] / 6 ] (you can do it the other way around, but the result should be the same)

Now you need to solve the aforementioned equation for k. They both have the same denominator, so you can scrap it. this should leave you with

[9 + sqrt(81 - 12k)] ~~/ 6~~ = 2 * [ [9 - sqrt(81 - 12k)] ~~/ 6~~ ]

which is

9 + sqrt(81 - 12k) = 2 * [9 - sqrt(81 - 12k)]

Now let's develop the right hand side of the equation, which is just multiplying by 2. This gives you:

9 + sqrt(81 - 12k) = 18 - 2 * sqrt(81 - 12k)

Now let's rearrange the equation by putting everything with k to the left hand side and everything NOT with k to the right. This gives you:

sqrt(81 - 12k) + 2 * sqrt(81 - 12k) = 18 - 9

Now it's just basic adding and subtracting. You should get:

3 * sqrt(81 - 12k) = 9

sqrt(81 - 12k) = 3

Now we get rid of the square root by elevating everything to the power of 2

[sqrt(81 - 12k)]^2 = 3^2

Which gives you

81-12k = 9

Now it should be smooth sailing from here. So solve it for k and you should get k=6. Plotting the graph of 3x^2 -9x + 6 will give you x = 1 and x = 2 as the roots, which fits as a possible answer. Despite this however, note that this solution might not be the best or most accurate. I am no expert and this is just a way I came up with to solve this problem and it yielded a possible answer. Does not mean it is 100% right.

laminutederire · July 4, 2016

In a nutshell, divide by 3 ( because morally, every polynomial function has its dominant degree with a factor ) then find what can be put in the form (x+a)^2 with its first coefficient being one, and the second being 3 (2a=3), then k/3 is equal to a^2 (k=3* (3/2)^2= 27/4 just like the other answer)

Dobbsjr · July 5, 2016

@Tuurahk @laminutederire

I actually found a much simpler solution. Idk if people outside of Korea actually learn this(tried googling it, but no results), but they make us memorize these weird equations. For example, (a + b)^2 = a^2 + 2ab + b^2 etc. So one of the equations looks like this: (x + a)(x + b). Basically the root of this quadratic is whatever makes the value of a or b zero. For example in the equation x^2 - 11 + 24 = 0 can be factored into (x - 3)(x- 8) = 0. You can easily find the roots here. X = 3 or 8.

Back to the original problem. First you would divide the original equation by 3 so x^2 - 3x + k/3 = 0. Then set the two roots as a and 2a(one is twice as big) and then put them into the equation above so (x - a)(x - 2a). Simplify that into x^2 - 3ax + 2a^2. Compare that with the original x^2 - 3x + k/3 and you can see that a = 1 and 2 = k/3. Therefore k = 6.

I know I explained a bit badly , so feel free to ask any questions.

vanished · July 5, 2016

11 minutes ago, Dobbsjr said:

@Tuurahk @laminutederire

I actually found a much simpler solution. Idk if people outside of Korea actually learn this(tried googling it, but no results), but they make us memorize these weird equations. For example, (a + b)^2 = a^2 + 2ab + b^2 etc. So one of the equations looks like this: (x + a)(x + b). Basically the root of this quadratic is whatever makes the value of a or b zero. For example in the equation x^2 - 11 + 24 = 0 can be factored into (x - 3)(x- 8) = 0. You can easily find the roots here. X = 3 or 8.

Back to the original problem. First you would divide the original equation by 3 so x^2 - 3x + k/3 = 0. Then set the two roots as a and 2a(one is twice as big) and then put them into the equation above so (x - a)(x - 2a). Simplify that into x^2 - 3ax + 2a^2. Compare that with the original x^2 - 3x + k/3 and you can see that a = 1 and 2 = k/3. Therefore k = 6.

I know I explained a bit badly , so feel free to ask any questions.

Yup, looks like it:

laminutederire · July 5, 2016

51 minutes ago, Dobbsjr said:

@Tuurahk @laminutederire

I actually found a much simpler solution. Idk if people outside of Korea actually learn this(tried googling it, but no results), but they make us memorize these weird equations. For example, (a + b)^2 = a^2 + 2ab + b^2 etc. So one of the equations looks like this: (x + a)(x + b). Basically the root of this quadratic is whatever makes the value of a or b zero. For example in the equation x^2 - 11 + 24 = 0 can be factored into (x - 3)(x- 8) = 0. You can easily find the roots here. X = 3 or 8.

Back to the original problem. First you would divide the original equation by 3 so x^2 - 3x + k/3 = 0. Then set the two roots as a and 2a(one is twice as big) and then put them into the equation above so (x - a)(x - 2a). Simplify that into x^2 - 3ax + 2a^2. Compare that with the original x^2 - 3x + k/3 and you can see that a = 1 and 2 = k/3. Therefore k = 6.

I know I explained a bit badly , so feel free to ask any questions.