get the roots of a function with substitution

[silent kite]

we have f(x)=-0.25 x^(4)+2.25 x^(2)+x-3
and i need to get the root of it
and i want to do it with substitution

i did x^2 = z

and then substituted it which meant -0.25z^2+2.25z then we had x-3 and i didnt really know how to do it there but my idea would be that x= squareroot(x^2) so for x we would substitute with squaroot(z) myb? im really confused on how to do it here

[unique owl]

multiply it by 4 and write as -x^2(x^2-9)+4(x-3). So, you factor out (x-3).

[silent kite]

wait but why -x^2 the orginal function is ^4

[unique owl]

-x² is multiplied by (x²-9). So, when you expand it gets -x⁴+...

[silent kite]

i see, but i dont get what u did at the begining like i mean in which way does this help me with the substitution, if i would try to substitute i would still have the issue that i cant substitute the x-3 or is substitution not possible in this example and i need to do it with polynomal division

[unique owl]

I don't see why you are so focused on substitution, which I think is unnecessary in this case. You just factor the polynomial as the linear factors are visible at once.

[silent kite]

i wanted to train it cuz it will come as an example in the exam

like its guaranteed that it will come

[slim imp]

Substitution of z = x^2 is a good idea if all the powers of x are even... it is more of a problem when we have the odd power x^1.

You have $f(x) = -\frac{1}{4}x^4 + \frac{9}{4}x^2 + x - 3 = 0$

$-\frac{x^2}{4}\left(x^2 - 9\right) + x - 3 = 0$

$-\frac{x^2}{4}(x + 3)(x - 3) + x - 3 = 0$

$-x^2(x + 3)(x - 3) + 4(x - 3) = 0$

$(x - 3)\left(4 - x^2(x+3)\right] = 0$

$(x - 3)\left(4 - 3x^2 - x^3\right) = 0$

$(x - 3)\left(4 - 4x^2 + x^2 - x^3\right) = 0$

$(x - 3)\left[4\left(1 - x^2\right) + x^2\left(1 - x\right)\right] = 0$

$(x - 3)\left[4(1 - x)(1 + x) + x^2(1 - x)\right] = 0$

$(x - 3)(1 - x)\left[4(1 + x) + x^2\right] = 0$

$(x - 3)(1 - x)\left(4 + 4x + x^2\right) = 0$

$(x - 3)(1 - x)(x + 2)^2 = 0$

[fathom flame]

Jay

[silent kite]

ah ok ok, i see now that explains it thanks!

.solved