could any1 help on this plz (1 Viewer)

[Sirius Black]

If you factorise x⁴-20x²+16=(x²+4x-4)(x²-4x-4)
OR=(x²+2sqrt7 x+4)(x²-2sqrt7 x+4) by using the result of x⁴+Ax²+B=(x²+ax+b)(x²-ax+b)

but the 4 roots to f(x)=x⁴-20x²+16=0 come from the second way of factorisation. Could any1 explain why the first one has to be discarded?

thanx

 

[icycloud]

x^4-20x^2+16 does not equal to (x^2+4x-4)(x^2-4x-4)

 

[Sirius Black]

x^4-20x^2+16 does not equal to (x^2+4x-4)(x^2-4x-4)

but the problem is that if u use the substituition given in the question u will get two sets of results
as A=2b-a² and B=b² if B=16 as in the question, u can hav b=+/-4 right? my question is that why the given result doesn't hold when b=-4 (LHS=/=RHS as u said)

 

[.ben]

is this from terry lee?

 

[Sirius Black]

is this from terry lee?

NO

could anyone help plz?

 

[insert-username]

(x²+4x-4)(x²-4x-4)

= x⁴ - 4x³ - 4x² + 4x³ - 16x² - 16x - 4x² + 16x +16

= x⁴ - 24x² + 16.

Which does not equal x⁴-20x²+16.

Therefore you get two different results because the substitution given to you is wrong. Or, your first result has to be discarded because it's entirely wrong.


Just quickly, to solve the equation, letting x² = a:

a² - 20a + 16

Therefore a = 10 ± 2√21 (quadratic formula)

Therefore x = ± √(10 ± 2√21). I'm pretty sure that's right.


I_F

 

[Sirius Black]

Therefore you get two different results because the substitution given to you is wrong. Or, your first result has to be discarded because it's entirely wrong.

hehe,ur reasoning sounds a bit like fudging... can u elaborate on that ?
If u take time to expand the RHS of the given formula, u can derive an expression in the form of LHS i.e. the result given in the question in NOT wrong.

Just quickly, to solve the equation, letting x² = a:

a² - 20a + 16

Therefore a = 10 ± 2√21 (quadratic formula)

Therefore x = ± √(10 ± 2√21). I'm pretty sure that's right.


I_F

this can be further simplifed to +/-√7 +/-√3

 

[Stan..]

If you factorise x⁴-20x²+16=(x²+4x-4)(x²-4x-4)
OR=(x²+2sqrt7 x+4)(x²-2sqrt7 x+4) by using the result of x⁴+Ax²+B=(x²+ax+b)(x²-ax+b)

but the 4 roots to f(x)=x⁴-20x²+16=0 come from the second way of factorisation. Could any1 explain why the first one has to be discarded?

thanx

Well, I agree with Icycloud that the First solution didn't come out.
The reason why the second works is cause it is a quadratic identity, and by splitting it into a form where it is factorised fully the roots are easy to determine. Just like partial fractions, easier to work with.

 

[icycloud]

The substitution does work, you just substituted it wrong.

Observe:

A = -20, B = 16

Thus, b = +/- 4

When b = 4, a = +/- 2Sqrt[7]
When b = -4, a = +/- 2Sqrt[3]

Therefore,
x^4-20x^2+16
= (x^2 + 2Sqrt[7]x + 4)(x^2 - 2Sqrt[7]x + 4) #
= (x^2 + 2Sqrt[3]x - 4)(x^2 - 2Sqrt[3]x - 4) #

Both versions work. Nowhere does (x^2+4x-4)(x^2-4x-4) come in...