Quadratic Equations Help

[WRXgearhead]

Hello all,

I having some trouble with a particular equation in my pre-calculus review and I am starting to get very frustrated. The instructions are "Find all real solutions". The equation is as follows:

(2-y)^4=3(2-y)^2+1

From what I remember, I need to move everything to one side of the equation, so that it equals 0. So then I get:

3(2-y)^2-(2-y)^4+1=0

I think that I have to distribute first, then do the powers. After distributing I get:

(6-3y)^2(-2-+y)^4+1

Then FOIL:

y^3-2y^2-2y^2-2y^2+9y^2-18y-18y+4y+4y+4y+36+1-8=0

Which seems really drawn out and long. After adding like terms, I get the following, which isn't a quadratic equation:

y^3+3y^2+24y+29=0

I am just very confused and at my wits end with equation any help would be very appreciated.

 

[stapel]

I having some trouble with a particular equation....:

(2-y)^4=3(2-y)^2+1

From what I remember, I need to move everything to one side of the equation, so that it equals 0. So then I get:

3(2-y)^2-(2-y)^4+1=0

Yes, though I would move things the other way 'round.

I think that I have to distribute first, then do the powers. After distributing I get:

(6-3y)^2(-2-+y)^4+1

I'm sorry, but I don't understand what happened here...?

Instead, try working directly with the quadratic-type form:

. . . . .(2 - y)⁴ = 3(2 - y)² + 1

. . . . .(2 - y)⁴ - 3(2 - y)² - 1 = 0

. . . . .[(2 - y)²]² - 3[(2 - y)²] - 1 = 0

What would you do with "x² - 3x - 1 = 0"? Probably use the Quadratic Formula to solve. So do that here:

. . . . .\(\displaystyle (2\, -\, y)\, =\, \dfrac{-(-3)\, \pm\, \sqrt{(-3)^2\, -\, 4(1)(-1)\,}}{2(1)}\)

. . . . .\(\displaystyle 2\, -\, y\, =\, \dfrac{3\, \pm\, \sqrt{13}\,}{2}\)

Then:

. . . . .\(\displaystyle 2\, -\, \dfrac{3\, \pm\, \sqrt{13}\,}{2}\, =\, y\)

...and so forth.

 

[HallsofIvy]

Hello all,

I having some trouble with a particular equation in my pre-calculus review and I am starting to get very frustrated. The instructions are "Find all real solutions". The equation is as follows:

(2-y)^4=3(2-y)^2+1

From what I remember, I need to move everything to one side of the equation, so that it equals 0. So then I get:

3(2-y)^2-(2-y)^4+1=0

Okay, though (2- y)^4- 3(2- y)^2- 1= 0 would be "standard form" with highest power first.

I think that I have to distribute first, then do the powers. After distributing I get:

(6-3y)^2(-2-+y)^4+1

No, this is wrong. 3(2- y)^2 is equal to 3(4- 4y+ y^2)= 12- 12y+ 3y^2 while (6- 3y)^2= 36- 36y+ 9y. Putting the 3 inside the square means it get squared also.

Then FOIL:

y^3-2y^2-2y^2-2y^2+9y^2-18y-18y+4y+4y+4y+36+1-8=0

This is also wrong, even for what you have above. It should be obvious, because of that fourth power, that you must have a y^4 in your equation but you don't. (6- 3y)^2+ (-2- y)^4+ 1= (36- 36+ 9y^2)+ (y^4+ 8y^3+ 24y^2+ 32y+ 16)+ 1= 0. However, as I said, you have the wrong equation.

Which seems really drawn out and long. After adding like terms, I get the following, which isn't a quadratic equation:

y^3+3y^2+24y+29=0

I am just very confused and at my wits end with equation any help would be very appreciated.

Did you not notice that the square and fourth power are of the same expression, 2- y? Did you not notice that you have only even powers? Let x= (2- y)^2 and the equation becomes x^2= 3x- 1 or x^2- 3x+ 1= 0. Solve that quadratic equation for x then solve (2- y)^2= x (Do NOT multiply that out- just take the square root of both sides.) In general a fourth degree equation has four roots.

 

[Steven G]

Yes, though I would move things the other way 'round.


I'm sorry, but I don't understand what happened here...?

Instead, try working directly with the quadratic-type form:

. . . . .(2 - y)⁴ = 3(2 - y)² + 1

. . . . .(2 - y)⁴ - 3(2 - y)² - 1 = 0

. . . . .[(2 - y)²]² - 3[(2 - y)²] - 1 = 0

What would you do with "x² - 3x - 1 = 0"? Probably use the Quadratic Formula to solve. So do that here:

. . . . .\(\displaystyle (2\, -\, y)\, =\, \dfrac{-(-3)\, \pm\, \sqrt{(-3)^2\, -\, 4(1)(-1)\,}}{2(1)}\)

. . . . .\(\displaystyle 2\, -\, y\, =\, \dfrac{3\, \pm\, \sqrt{13}\,}{2}\)

Then:

. . . . .\(\displaystyle 2\, -\, \dfrac{3\, \pm\, \sqrt{13}\,}{2}\, =\, y\)

...and so forth.

I am sorry for the above post. The poster is incorrectly saying that this is a quadratic equation in (2 - y) while it is a quadratic equation in (2 - y)² . So [FONT=MathJax_Math](2 - y)² [/FONT][FONT=MathJax_Main]=[3 +/- sqrt(13)]/2[/FONT]

The cleanest way to do such a problem, in my opinion, is to do almost as Dennis pointed out and let k = (2 - y)²

 

[Steven G]

I thought you promised never to read any of my posts :twisted:

Denis, I guess that you do not know me too well as I never take my own advise. After all, I may be stupid but I am not dumb.