Question

9x^4+ 20x^2 + 12 in quadratic form

Answer 1

Answer:

A "pure" quadratic has the form:

ax^2 + bx + c

To put this in broad, general (and crude) terms, a quadratic is:

"Some number, 'a', times a perfect square plus some number. 'b', times whatever-is being squared-next to-the-a (x) plus some other number, 'c'"

So can we make 9x^4+20x^2+12 fit this pattern? Let's look at this expression term-by-term:

9x^4

Can this be expressed as "Some number, 'a', times a perfect square"? Answer: Yes, in two ways:

9x^4 = 9*(x^2)^2

or

1*(3x^2)^2

Next

10x^2

Can this be expressed as "some number. 'b', times whatever-is being squared-next to-the-a"? The answer depends on which of the expressions we use for 9x^4.

If we use 9*(x^2)^2 then "whatever-is being squared-next to-the-a" would be x^2. Can we express 20x^2 as some number times x^2? Obviously yes: 20*x^2.

If we try to use 1*(3x^2)^2 for the 9x^4 then "whatever-is being squared-next to-the-a" would be 3x^2. Can we express 20x^2 as some number times 3x^2? Although not very obvious, the answer is yes: 20x^2 = (20/3)*3x^2

And of course 12 can be our 'c'. So there are two ways to express 9x^4+20x^2+12 in quadratic form:

9x^4+20x^2+12 = 9*(x^2)^2 + 20*x^2 + 12 with the a = 9, b = 20 and the c = 12

or

9x^4+20x^2+12 = 1*(3x^2)^2 + (20/3)*3x^2 + 12 with the a = 1, b = 20/3 and the c = 12.

If you really want these to look like a quadratic, then you can use a temporary variable. Make the temporary variable equal to "whatever-is being squared-next to-the-a". So for 9*(x^2)^2 + 20*x^2 + 12, let q = x^2. This makes the expression: 9q^2 + 20q + 12. For 1*(3x^2)^2 + (20/3)*3x^2 + 12, let q = 3x^2. This m

Step-by-step explanation: