# Quick help needed on a simple problem (not simple4me)

1. Mar 20, 2005

### 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 im stuck! i dont know if i should go back and rewrite z^4 as $$z^2z^2$$ or what??

any help is appreciated!! thanks!!

2. Mar 20, 2005

### 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??

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Last edited: Mar 20, 2005
3. Mar 20, 2005

### mathzeroh

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

4. Mar 20, 2005

### 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 dont work though... ??

5. Mar 20, 2005

### Data

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

6. Mar 20, 2005

### mathzeroh

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

7. Mar 20, 2005

### xanthym

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

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Last edited: Mar 20, 2005
8. Mar 20, 2005

### 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!! :) :)

9. Mar 20, 2005

### mathzeroh

x = 25 OR x = 3

is that right???

10. Mar 20, 2005

### 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

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Last edited: Mar 20, 2005
11. Mar 20, 2005

### 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!!