Quadratic: (t - 2)^2 = (2 + 2)^2 + 64 (Me vs. Answer Key)

[WaterOtter]

HELP!! Any way I do this problem, I come up with no real solution. The Answer Key says x=-8. What am I doing wrong???

(t-2)^2=(2+2)^2+64

Test is tonight!!

 

[stapel]

The Answer Key says x=-8.

Since the quadratic is in "t", not "x", the answer key is obviously incorrect.

What am I doing wrong?

Since you haven't shown your work, I'm afraid there is no way to tell. Sorry!

Eliz.

 

[WaterOtter]

Sorry, my mistake: Anwer Key t=+8

(t-2)^2=(t+2)^2+64
(t-2)^2-(t+2)^2=64
Sqrt((t-2)^2-(t+2)^2)=sqrt(64)
t-2-t+2=8
t-t-2+2=8
0=8 no solution

 

[masters]

Assuming in your original equation, you meant (t - 2)[sup:1bzwfnlw]2[/sup:1bzwfnlw] = (t + 2)[sup:1bzwfnlw]2[/sup:1bzwfnlw] + 64 you must square each of the two binomials and combine terms:

t[sup:1bzwfnlw]2[/sup:1bzwfnlw] - 4t + 4 = t[sup:1bzwfnlw]2[/sup:1bzwfnlw] + 4t +4 + 64

Combining terms, we get:

-8t = 64

And dividing by -8:

t = -8

 

[WaterOtter]

Ah Ha! Just square the binomials...I get it now. THANK YOU!!

 

[stapel]

(t-2)^2=(t+2)^2+64
(t-2)^2-(t+2)^2=64
Sqrt((t-2)^2-(t+2)^2)=sqrt(64)
t-2-t+2=8 <= NO

They should have mentioned, back when you were introduced to radicals, that roots are not distributive. The square root of a sum or difference is NOT the sum or difference of the square roots! :shock:

. . . . .\(\displaystyle \sqrt{3^2\, + \,4^2}\, =\, \sqrt{9\, +\, 16}\, =\, \sqrt{25}\, = 5\)

...but:

. . . . .\(\displaystyle \sqrt{3^2}\, +\, \sqrt{4^2}\, =\, 3\, +\, 4\, =\, 7\)

Eliz.