Equations in quadratic form: don't understand example

[jenjen]

There is an example in my book which has the following problem. The answer is already provided and I can solve half the problem... however I dont quite understand it towards the end
this is exactly how its done in the book.

this is the equation.
(x^2-1)^2 + (x^2-1)-12=0

so let U= x^2-1 so that u^2= (x^2-1)^2

The orginal equation becomes
(u^2-1)^2 + (u^2-1)-12=0
(ut4) (u-3)=0
U=-4 or U=3

solve for x
because u=x^2-1 we have

x^2-1=-4 or x^2-1=3
x^2=-3 and x^2 =4

the second solution set ends up being
x=-2 and x= 2

^^^^^^^^^^^this is the part i do not understand.

 

[wjm11]

Re: EQUATIONS THAT ARE IN QUADRATIC FORM

x^2=-3 and x^2 =4

the second solution set ends up being
x=-2 and x= 2

Hi, Jenjen,

The first equation, x^2=-3, has no Real solution (only imaginary/complex number solutions).

The second equation, x^2 =4, has two Real solutions for x: +2 and –2. Both those numbers, when squared, will equal 4.

Additionally, both those numbers can be “plugged in” to the original equation,

(x^2-1)^2 + (x^2-1)-12=0

to check for accuracy.

 

[stapel]

the second solution set ends up being
x=-2 and x= 2

^^^^^^^^^^^this is the part i do not understand.

You have:

. . . . .\(\displaystyle x^2\, =\, 4\)

You can either take the square root of each side, putting a "plus-minus" on the non-variable side:

. . . . .\(\displaystyle \sqrt{x^2}\, =\, \pm \sqrt{4}\)

. . . . .\(\displaystyle x\, =\, \pm 2\)

...or you can use factoring:

. . . . .\(\displaystyle x^2\, -\, 4\, =\, 0\)

. . . . .\(\displaystyle (x\, -\, 2)(x\, +\, 2)\, =\, 0\)

. . . . .\(\displaystyle x\, -\, 2\, =\, 0\, \, \mbox{ or }\, \, x\, +\, 2\, =\, 0\)

. . . . .\(\displaystyle x\, =\, 2\, \, \mbox{ or } \, \, x\, =\, -2\)

Either way, you'll end up with x = -2, +2.

Eliz.

 

[jenjen]

How come I cant use the -3 and factor.

 

[Deleted member 4993]

How come I cant use the -3 and factor.

Because, in that case (x^2 + 3 =0), you cannot find a "real" number "that when squared" (x^2), will give a negative number.

The books answer assumes that you have not been exposed to "complex" numbers.

 

[jenjen]

Thank you!!

 

[stapel]

How come I cant use the -3 and factor.

I showed the factorization with the "x[sup:2och78y8]2[/sup:2och78y8] = 4" using the difference-of-squares formula that you've memorized. What formula do you have for factoring a sum of squares? (Hint: There isn't one.)

Eliz.