How do you solve a quadratic equation via substitution?

Question

I'm in desperate need of help with this. I was reading through my text book when it read:

Solve x^2^3 + x^1^3 - 6 =0

The book gave me the answer but I'm confused about how they got it.

The book continues by saying:

Substituting y=x^1^3 gives

y^2 + y -6 = 0 (HOW DID YOU GET THAT???)

...

(y+3)(y-2) = 0 (AND THIS!!!)

Either y+3=0 OR y-2=0, so y = -3 or y = 2

The eventual answer was x = -27 or x = 8 which I dont know how it was found

Can someone tell me how to do this please!!!

Answer 1

This technique/strategy is used when the degree of the equation you are asked to solve is higher than the second degree (quadratic), but the form (pattern) of the equation is like a quadratic.  That is, it has three terms (usually), the exponent of the variable(s) in the first term is twice the exponent of the variable(s) in the second term, and the second term is a composite of the factors(s) from the first and third terms.   (Without getting too complicated, the third term could contain variable(s).  This would change the structure, but not the pattern.)

Your equation is an example.

I don't particularly like the way it is written here.  Let me rewrite it to make this clearer.

(x^3)^2 + (x^3) - 6 = 0

Do you see that this is the same as what you wrote ?

Now, if you let y = x^3 (temporarily), then you can write the equation as:

y^2 + y -6 =0

That is a familiar quadratic, yes ?

factor and solve for 'y'

(y + 3)(y - 2) = 0

y = -3 and y = 2

But remeber, y = (x^3)  (that was our substitution)

We were asked to solve for x

x = -27 and x = 8

Is that clear ?  Good.  Try to do a few more from your book.  Good luck.

Answer 2

You're either missing another eq or you're missing the y variable in your example.  Most likely you're missing an equation because in order to do substitution, you need 2 eqs.  You solve one for one of the variables and then insert it into the 2nd eq.

I think I see what they're doing.  They are assigning x^1^3 the value of y.  So in the eq (x^2)^3 + (x^1)^3 - 6=0, replace x^1^3 with y

x^2^3 + y - 6.  Now, (x^2)^3 = ((x^1)^3)^2, so you have

y^2 + y -6, then they just factored

(y+3)(y-2)= 0 so either y+3=0 OR y-2=0, so y = -3 or y = 2

Not sure where they get the x values though... to me, it seems now that since y = 2, then x^1^3=2, so it would be the cube rt of 2, but they're saying that it's 2^3.  Maybe someone else can explain that part.