• Jun 20th 2013, 09:44 PM
Mathnood768
http://puu.sh/3kEIh.png

My attempt: I wrote it down on paper and I have nothing I could use to take a picture of, I'm not sure how I can write it out but:

square both sides to get rid of the radical sign on the left side then use FOIL on right side.

q^2/2 + 11 = q^2 - 2q + 2

multiply 2

q^2 - 4q - 18

then i put it into the quadratic formula, q^2 = a -4 = b -18 = c

after that, I ended up with http://puu.sh/3kEO7.png

the answer for this question is suppose to be q = 2 + 2(sqrt(6))

i don't know how to get the right answer.
• Jun 20th 2013, 09:53 PM
ibdutt
On checking we find tat for this equation we will not get the answer given. please check if the question has been correctly pasted.
• Jun 20th 2013, 10:13 PM
Mathnood768
yes, i pasted my question correctly.
• Jun 20th 2013, 10:43 PM
ibdutt
• Jun 21st 2013, 11:21 AM
Mathnood768
http://puu.sh/3kEIh.png

this question solved is suppose to be q = 2 + 2(sqrt(6)) {apparently that is the correct answer}

but I get http://puu.sh/3kEO7.png
• Jun 21st 2013, 02:58 PM
Soroban
Hello, Mathnood768!

You're making a number of blunders . . .

Quote:

$\sqrt{\frac{q^2}{2} + 11} \:=\:q-1$

Square both sides: . $\left(\sqrt{\frac{q^2}{2} + 11}}\right)^2 \;=\;(q-1)^2$

. . . . . . . . . . . . . . . . . . . $\frac{q^2}{2} + 11 \;=\;q^2-2q + 1$

Multiply by 2: . . . . . . . . . $q^2 + 22 \;=\;2q^2 - 4q + 2$

. . . . . . . . . . . . . . . . $q^2 - 4q - 20 \;=\;0$

Quadratic Formula: . $q \;=\;\frac{\text{-}(\text{-}4) \pm\sqrt{(\text{-}4)^2 - 4(1)(\text{-}20)}}{2(1)} \;=\;\frac{4\pm\sqrt{96}}{2}$

. . . . . . . . . . . . . . . $q \;=\;\frac{4\pm 4\sqrt{6}}{2} \;=\;\frac{2(2\pm2\sqrt{6})}{2} \;=\;2 \pm2\sqrt{6}$

But we find that. $q \:=\:2-2\sqrt{6}$ndoes not check.
. . It is an extraneous root.

The only root is: . $q \:=\:2 + 2\sqrt{6}$