Quick help needed on a simple problem (not simple4me)

[mathzeroh]

quick help needed on a "simple" problem (not simple4me)

hey everyone i hope ur all doing great! ok tomorrows my math final and i have a slight dilemma on the review sheet and i would appreciate it a lot if someone helped me with it!


here it is:

z^4+75=28z^2

ok so i first brought the 75 to the other side which gave me:

z^4=28z^2-75

and now I am stuck! i don't know if i should go back and rewrite z^4 as z^2z^2 or what??

any help is appreciated! thanks!

 

[xanthym]

Hint: Make the substitution {x = z²} and solve the resulting quadratic equation with "x" for the variable (for the first step). Can you see what to do after that??


~~

 

[mathzeroh]

Hint: Make the substitution {x = z²} and solve the resulting quadratic equation with "x" for the variable (for the first step). Can you see what to do after that??


~~

let me try and see if i understood what u meant! i'll be back in a bit! thanks btw

 

[mathzeroh]

ok so i took ur advice and if i understood it correctly this is what i did with it, however i checked my answer and it didn't check...

z^4+75=28z^2
z^4+75=28x; x=z^2
z^4=28x-75
x^2=28x-75, x=z^2 so x^2=z^4
x^2-28x-75=0

then i plugged this into the quadtratic and got this:

x=14\pm\sqrt{271}

so i got this approx.:

x\approx30.460​

OR​

x\approx-2.460​

but those answers don't work though... ??

 

[Data]

Should that really be x^2 - 28x - 75 =0? I like positive 75s more...

 

[mathzeroh]

but does that make a difference when it comes to the quadtratic formula?

 

[xanthym]

but does that make a difference when it comes to the quadtratic formula?

YES! The correct equation is given below. Can you see how to factor it??
(Or you can also use the quadratic formula.)
x² - 28*x + 75 = 0


~~

 

[mathzeroh]

holy crap ur right! when i brought the 28x over, i should've ADDED a negative -75 to both sides, which would make it +75 yep ur right!

brb! :) :)

 

[mathzeroh]

x = 25 OR x = 3


is that right?

 

[xanthym]

CORRECT!
Remember, solving the equation in "x" is just the first step. You then must place the "x" solutions back into your original {x = z²} substitution equation and solve for "z" (for both "x" solutions).
x² - 28*x + 75 = 0

~~

 

[mathzeroh]

yea i went back and did that but it wasn't that obvious for me at first :D

ok so in the solution of x=3, since z^2=x, that means that z=the square root of x, or in this case, the square root of 3.

and i plugged that value in the original formula of z^4+75=28z^2 and it worked.

and i did the same thing for the other value of x, which was 25 and did the same thing!

so basically, z=/sqrt(3) or z=5 (because the sqrt of 25=5)

thanks a lot man!

p.s. both of those are plus or minus btw!