We know that
A difference of two perfect squares (A² - B²) <span>can be factored into </span><span> (A+B) • (A-B)
</span> then
x ^4-4--------> (x²-2)*(x²+2)
(x²-2)--------> (x-√2)*(x+√2)
x1=+√2
x2=-√2
the other term
(x²+2)=0-> x²=-2-------------- x=(+-)√-2
i <span> is called the </span><span>imaginary unit. </span><span>It satisfies </span><span> i</span>²<span> =-1
</span><span>Both </span><span> i </span><span> and </span><span> -i </span><span> are the square roots of </span><span> -1
</span><span>√<span> -2 </span></span> =√<span> -1• 2 </span><span> = </span>√ -1 •√<span> 2 </span> =i • <span> √<span> 2 </span></span>
The equation has no real solutions. It has 2 imaginary, or complex solutions.
x3= 0 + √2<span> <span>i
</span></span>x4= 0 - √2<span> i </span>
the answer is
the values of x are
x1=+√2
x2=-√2
x3= 0 + √2 i
x4= 0 - √2 i
Answer:
1) -32+12x 2) 2(3x-4)
Step-by-step explanation:
I hope this is right!
The answer would be 90 because all you do is add the numbers ! (: pls mark me brainliest
Answer:
p ∈ IR - {6}
Step-by-step explanation:
The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2
is all R2 ⇔
And also u and v must be linearly independent.
In order to achieve the final condition, we can make a matrix that belongs to 
using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.
Let's make the matrix :
![A=\left[\begin{array}{cc}3&1&p&2\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%261%26p%262%5Cend%7Barray%7D%5Cright%5D)
We used the first vector ''u'' as the first column of the matrix A
We used the second vector ''v'' as the second column of the matrix A
The determinant of the matrix ''A'' is
We need this determinant to be different to zero
The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that
We can write : p ∈ IR - {6}
Notice that is 
⇒
If we write 
, the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.
