What is (5i-j+2k)^2 ?

The solution to (5i-j+2k)^2 is (j^2-4jk+4k^2-25)+(-10j+20k)i

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simplify (5i-j+2k)^2

simplify

(5 i - j + 2 k)^{2}

(j^{2} - 4 jk + 4 k^{2} - 25) + (- 10 j + 20 k) i

Solution steps

(5 i - j + 2 k)^{2}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

(5 i - j + 2 k)^{2} = (5 i - j + 2 k) (5 i - j + 2 k)

= (5 i - j + 2 k) (5 i - j + 2 k)

Distribute parentheses

= 5 i 5 i + 5 i (- j) + 5 i 2 k - j \cdot 5 i - j (- j) - j \cdot 2 k + 2 k \cdot 5 i + 2 k (- j) + 2 k \cdot 2 k

Simplify

5 i 5 i + 5 i (- j) + 5 i 2 k - j \cdot 5 i - j (- j) - j \cdot 2 k + 2 k \cdot 5 i + 2 k (- j) + 2 k \cdot 2 k : j^{2} - 10 ji + 20 ki - 4 jk + 4 k^{2} - 25

= j^{2} - 10 ji + 20 ki - 4 jk + 4 k^{2} - 25

Rewrite in standard complex form:

(j^{2} - 4 jk + 4 k^{2} - 25) + (- 10 j + 20 k) i

= (j^{2} - 4 jk + 4 k^{2} - 25) + (- 10 j + 20 k) i

(11 + 1 - x^{2}) (1 - 1 - x^{2})

\frac{sin ( o ) 2 α + cos ( 2 α ) + 1}{2 ( cos ( α ) + sin ( o ) α )}

\frac{( 3 ^{x} + 2 ^{x + 1} )}{6 ( 3 ^{x - 1} + 2 ^{x} )}

(10 - \frac{10 π}{x + 4})^{2}

(m + 7) (m - 7)

What is (5i-j+2k)^2 ?
The solution to (5i-j+2k)^2 is (j^2-4jk+4k^2-25)+(-10j+20k)i