How to start solving

[Mystic]

Hi,

I am new here and my native language is not english as I am from northern Europe. In our coyntry there is also different term used for algebra and calculus so I am not excactly sure if this question is now on the right forum.

But the question is related to solving functions higher than second degree. So the equation is like this:

So at first I thought I could just take square root from both and factor out zeros from second degree equation but I forgot root could be either (+) or (-). So you can't do like this if you don't do equation to both possibilities.

So what would be the best way to solve that? I have not yet learned how to solve functions higher than second degree so there should be some method of factoring out or something else I am not sure.

 

[Deleted member 4993]

Hi,

I am new here and my native language is not english as I am from northern Europe. In our coyntry there is also different term used for algebra and calculus so I am not excactly sure if this question is now on the right forum.

But the question is related to solving functions higher than second degree. So the equation is like this:

So at first I thought I could just take square root from both and factor out zeros from second degree equation but I forgot root could be either (+) or (-). So you can't do like this if you don't do equation to both possibilities.

So what would be the best way to solve that? I have not yet learned how to solve functions higher than second degree so there should be some method of factoring out or something else I am not sure.

There are many ways to solve it. But "taking square-root" is NOT the easiest path. A square-root can be either positive or negative.

One of the ways I could solve this:

....... ←...... 4x² - 9 = (2x-3)(2x+3)

(2x-3)² * [(2x+3)² - 1²] = 0

continue........

 

[JeffM]

You certainly can solve it using square roots, but this particular problem can be solved more easily.

[math](4x^2 - 9)^2 = (2x - 3)^2 \implies \{(2x + 3)(2x - 3)\}^2 = (2x - 3)^2 \implies\\ (2x + 3)^2(2x - 3)^2 = (2x - 3)^2.\\ \text {CASE I: } 2x - 3 = 0 \implies x = \dfrac{3}{2}.\\ \text {CASE II: } 2x - 3 \ne 0 \implies (2x + 3)^2 = 1 \implies\\ 4x^2 + 12x + 9 = 1 \implies x^2 + 3x + 2 = 0 \implies\\ (x + 2)(x + 1) = 0 \implies x = -1 \text { or } x = - 2.[/math]

 

[pka]

But the question is related to solving functions higher than second degree. So the equation is like this:

Chasing cases is not my cup of tea. So look here. Scroll down to the expanded form.

 

[Dr.Peterson]

Hi,

I am new here and my native language is not english as I am from northern Europe. In our coyntry there is also different term used for algebra and calculus so I am not excactly sure if this question is now on the right forum.

But the question is related to solving functions higher than second degree. So the equation is like this:

So at first I thought I could just take square root from both and factor out zeros from second degree equation but I forgot root could be either (+) or (-). So you can't do like this if you don't do equation to both possibilities.

So what would be the best way to solve that? I have not yet learned how to solve functions higher than second degree so there should be some method of factoring out or something else I am not sure.

Your method works just fine (though perhaps a little slower than the others). Just take two cases: [math]4x^2-9=2x-3[/math] or [math]4x^2-9=-(2x-3)[/math]
Solve both resulting quadratic equations, and you will get the three distinct roots (one of them twice).

 

[Mystic]

Thank you for your help! Nice.

 

[Deleted member 4993]

One of the ways I could solve this:

............................. ←....................... 4x2 - 9 = (2x-3)(2x+3)

(2x-3)² * [(2x+3)² - 1²] = 0

(2x-3)^2 * [(2x-3)+1] * [(2x-3)-1] = 0

(2x-3)^2 * (2x-2) * (2x-4) = 0 .............................Applying rule of factor of 0

2x-3 = 0 → x = 3/2 .............................(repeated root).......................... or

2x-2 = 0 → x = 1 .......................... or

2x-4 = 0 → x = 2

 

[Steven G]

(2x-3)^2 * [(2x-3)+1] * [(2x-3)-1] = 0

(2x-3)^2 * (2x-2) * (2x-4) = 0 .............................Applying rule of factor of 0

2x-3 = 0 → x = 3/2 .............................(repeated root).......................... or

2x-2 = 0 → x = 1 .......................... or

2x-4 = 0 → x = 2

Are you saying that x=1 and x=2 are solutions to the original equations? If you are then I am afraid that you're the only one saying this.
It's been a while, but I think it is corner time for you. Pick a dark one, please.

 

[Deleted member 4993]

Are you saying that x=1 and x=2 are solutions to the original equations? If you are then I am afraid that you're the only one saying this.
It's been a while, but I think it is corner time for you. Pick a dark one, please.

I broke my cardinal rule - being in a hurry I did not check answer. I am grateful that the corner has been kept clean by its former occupant.