What's the right way to solve g(x)=4x^2−28x+49? (not understanding video)

[bbm25]

The problem is because this video apparently teaches one thing but the following exercise does another.

You don't need to watch the whole video, just skip straight to 4:53 to see all his working.

https://www.khanacademy.org/math/al...jYTFmZjVmMzUzZjEMCxIIRmVlZGJhY2sYgICA8MjDgQoM

In this video, he divides everything in the equation by 4.

In the practice following this video, the following question was given:

g(x)=4x^2−28x+49

and here was the site's answer:

< Link to sexually-oriented image of young girl removed >

here was my answer:
(4x^2)/4 - 28x/4 + 49/4

x^2 -7x +49/4

(x-7/2)(x-7/2)
when i open the bracket

x^2 -7x/2 - 7x/2 +49/4

= x^2 -7x +49/4

= (x-7/2)^2

i tried to follow this video. From what i understand, the practice exercise, doesn't.. Please help? What should i actually do and why?

 

[tkhunny]

The problem is because this video apparently teaches one thing but the following exercise does another.

You don't need to watch the whole video, just skip straight to 4:53 to see all his working.

https://www.khanacademy.org/math/al...jYTFmZjVmMzUzZjEMCxIIRmVlZGJhY2sYgICA8MjDgQoM

In this video, he divides everything in the equation by 4.

In the practice following this video, the following question was given:

g(x)=4x^2−28x+49

and here was the site's answer: <objectionable link removed >

here was my answer:
(4x^2)/4 - 28x/4 + 49/4

x^2 -7x +49/4

(x-7/2)(x-7/2)
when i open the bracket

x^2 -7x/2 - 7x/2 +49/4

= x^2 -7x +49/4

= (x-7/2)^2

i tried to follow this video. From what i understand, the practice exercise, doesn't.. Please help? What should i actually do and why?

1) You cannot SOLVE a function definition. g(x)=4x^2−28x+49 -- That's it. We now know what g(x) represents.
2) We can solve g(x) = 0 or 4x^2 − 28x + 49 = 0

For #2, I would think this:
----- a) 4x^2 is a perfect square - (2x)^2
----- b) 49 is a perfect square - (7)^2
----- c) he coefficient on the linear term is negative - -28
----- d) Let's just try (2x-7)^2 and see if we get g(x). If this succeeds, we're well on our way to a solution.
----- e) This is essentially "Completing the Square". It's just really easy when the adjustment is zero (0).

You could also find some other way to factor g(x).
You could also use the quadratic formula.

 

[Dr.Peterson]

The problem is because this video apparently teaches one thing but the following exercise does another.

You don't need to watch the whole video, just skip straight to 4:53 to see all his working.

https://www.khanacademy.org/math/al...jYTFmZjVmMzUzZjEMCxIIRmVlZGJhY2sYgICA8MjDgQoM

In this video, he divides everything in the equation by 4.

In the practice following this video, the following question was given:

g(x)=4x^2−28x+49

and here was the site's answer: <objectionable link removed >

here was my answer:
(4x^2)/4 - 28x/4 + 49/4

x^2 -7x +49/4

(x-7/2)(x-7/2)
when i open the bracket

x^2 -7x/2 - 7x/2 +49/4

= x^2 -7x +49/4

= (x-7/2)^2

i tried to follow this video. From what i understand, the practice exercise, doesn't.. Please help? What should i actually do and why?

I agree with you that the exercise does not belong after that video. The video's problem is stated as

Complete the square to solve 4x^2 + 40x - 300 = 0
​

while the exercise is

Rewrite the function by completing the square: 4x^2 - 28x + 49.
​

These are closely related ideas, but are done in rather different ways in detail.

There should have been an example that showed how to rewrite a function where a is not 1. I only see one where a is 1; I can only guess that they figured the example they gave would supply ideas for doing this different task. But I don't think it is adequate.

There are several ways to carry out the task, which is definitely not easy if you haven't seen it before. A big difference between the two tasks is that you can divide an equation by 4 without changing the solutions, but you can't divide a function (expression) by 4 without changing its meaning. So rather than divide, we usually would factor out the 4. (As I said, there are other ways, too.)

Learn from their solution to this problem. And learn also that Khan Academy is not perfect. If you want, you might write to them and point out the difficulty caused by the lack of a worked example of this type.

 

[Steven G]

The problem is because this video apparently teaches one thing but the following exercise does another.

You don't need to watch the whole video, just skip straight to 4:53 to see all his working.

https://www.khanacademy.org/math/al...jYTFmZjVmMzUzZjEMCxIIRmVlZGJhY2sYgICA8MjDgQoM

In this video, he divides everything in the equation by 4.

In the practice following this video, the following question was given:

g(x)=4x^2−28x+49

and here was the site's answer: < objectionable link removed >

here was my answer:
(4x^2)/4 - 28x/4 + 49/4

x^2 -7x +49/4

(x-7/2)(x-7/2)
when i open the bracket

x^2 -7x/2 - 7x/2 +49/4

= x^2 -7x +49/4

= (x-7/2)^2

i tried to follow this video. From what i understand, the practice exercise, doesn't.. Please help? What should i actually do and why?

Your problem is that 4x^2−28x+49 DOES NOT EQUAL (4x^2)/4 - 28x/4 + 49/4.

Instead 4x^2−28x+49 DOES EQUAL 4{(4x^2)/4 - 28x/4 + 49/4} and this equals 4{ x^2 -7x +49/4} = 4(x-7/2)^2 = 2^2*(x-7/2)^2 = {2*(x-7/2)}^2 = (2x-7)^2.