Hey, why are these two the same value? I thought the left side can be simplified to
x^2 + 6x + 3
so i find no relation with the right side
Squaring on both side:
(\sqrt{x^4+6x^2+9})^2 = (x^2+3)^2
x^4+6x^2+9 = (x^2+3)^2
\begin{aligned} (x^2+3)^2&= (x^2)^2+3^2+2 \cdot(3)\cdot(x^2) \\ &=x^4+9+6x^2 \end{aligned}
but there are no brackets around x^2+3
watch this : https://www.youtube.com/watch?v=0gicD4STzpg
or for a deep dive , look at Khan Academy https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:eq/x2ec2f6f830c9fb89:sqrt-eq/v/solving-square-root-equations-w-two-solutions
i watched all the prepswift, I don’t see the relevance of these videos with respect to my question however. You write (x+3)^2 but that’s not what is given as option B or am I misunderstanding what you write. Thanks
That is how you solve the radical(Squaring both the sides) as show in my earlier comment or ,
There is other more complex way to solve this, and this how you can do it
\begin{aligned}
\sqrt{x^4+6x^2+9} &= (x^4+6x^2+9)^{\frac{1}{2}}\\
&=(x^4+3x^2+3x^2+9x)^{\frac{1}{2}}\\
&=(x^2(x^2+3)+3(x^2+3))^{\frac{1}{2}}\\
&=((x^2+3)(x^2+3))^{\frac{1}{2}} \\
&=((x^2+3)^2)^{\frac{1}{2}}\\
&=(x^2+3)
\end{aligned}
Hint: let t = x^2
is this the completing the square technique i saw on prepswift?
i don’t know what you mean with both sides, there is only one side
ah yes ok that way, but then are you allowed to connect both sides to eachother with an equal sign?? i never saw that in such a problem, that’s why i didn’t get it
The goal in a QC problem is to check whether the two sides are equal (or greater/less than/neither).
Also: \sqrt{x^4 + 6x^2 + 9} is not x^2 + 6x + 3