Question

PLEASE HELP ILL GIVE BRAINLIEST

Answer 1

1)

$(-2+\sqrt{-5})^2\implies (-2+\sqrt{-1\cdot 5})^2\implies (-2+\sqrt{-1}\sqrt{5})^2\implies (-2+i\sqrt{5})^2 \\\\\\ (-2+i\sqrt{5})(-2+i\sqrt{5})\implies +4-2i\sqrt{5}-2i\sqrt{5}+(i\sqrt{5})^2 \\\\\\ 4-4i\sqrt{5}+[i^2(\sqrt{5})^2]\implies 4-4i\sqrt{5}+[-1\cdot 5] \\\\\\ 4-4i\sqrt{5}-5\implies -1-4i\sqrt{5}$

3)

let's recall that the conjugate of any pair a + b is simply the same pair with a different sign, namely a - b and the reverse is also true, let's also recall that i² = -1.

$\cfrac{6-7i}{1-2i}\implies \stackrel{\textit{multiplying both sides by the denominator's conjugate}}{\cfrac{6-7i}{1-2i}\cdot \cfrac{1+2i}{1+2i}\implies \cfrac{(6-7i)(1+2i)}{\underset{\textit{difference of squares}}{(1-2i)(1+2i)}}} \\\\\\ \cfrac{(6-7i)(1+2i)}{1^2-(2i)^2}\implies \cfrac{6-12i-7i-14i^2}{1-(2^2i^2)}\implies \cfrac{6-19i-14(-1)}{1-[4(-1)]} \\\\\\ \cfrac{6-19i+14}{1-(-4)}\implies \cfrac{20-19i}{1+4}\implies \cfrac{20-19i}{5}\implies \cfrac{20}{5}-\cfrac{19i}{5}\implies 4-\cfrac{19i}{5}$

Answer 2

Answer:

1) -1-4√(5)*i

2)(8/41)+(10/41)i

3)4+i

4)x=(1/3)±[√(23)/3]i OR x=(1/3)+[√(23)/3]i and x=(1/3)-[√(23)/3]i

Step-by-step explanation:

1) remember the special product:

(a+b)²=a²+2ab+b²

in this situation, a=-2 and b=√(-5)

(-2+√(-5))²=(-2)²+2(-2)(√(-5))+(√(-5))²

                =4-4*√(-5)+(-5)

                =-1-4√(-5)

√(-5)=-√(5)*i, so the final answer is: -1-4√(5)*i

2) remember the special product:

(a+b)(a-b)=a²-b²

multiply by the conjugate of the denominator (so multiply the numerator and denominator by 4+5i to remove i from the denominator, I'll demonstrate)

  2/(4-5i) * (4+5i)/(4+5i)

=2(4+5i)/[(4-5i)(4+5i)]

use the special product; in this situation, a=4 and b=5i

=(8+10i)/[16-(5i)²]

simplify: (5i)²=5²*(√(-1))²=25*-1=-25

(8+10i)/[16-(5i)²]=(8+10i)/[16-(-25)]

                        =(8+10i)/41

and finally, divide to get the answer in the correct form:

                        =(8/41)+(10/41)i

you can use a calculator to write 8/41 and 10/41 as decimals instead.

3) again, multiply by the conjugate of the denominator to both the top and the bottom:

 (6-7i)/(1-2i) * (1+2i)/(1+2i)

=(6-7i)(1+2i)/[(1-2i)(1+2i)]

now for the multiplication of the numerator:

(6-7i)(1+2i)=6(1+2i)-7i(1+2i)=(6+12i)-(7i+14i²)

i²=-1, so 14i²=-14

(6+12i)-(7i+14i²)=6+12i-7i-(-14)=6+5i+14=20+5i

as for the denominator, do what we did in 2) and use the special product to get:

(1-2i)(1+2i)=1-(2i)²=1-(-4)=5

so the answer is (20+5i)/5

now divide to convert the answer to the proper form/simplify

(20/5)+(5/5)i=4+i

4) use the quadratic formula, this is how you solve second-degree equations that aren't very easy to factor.

the formula: for ax^2+bx+c=0, x=[-b±√(b²-4ac)]/2a

in this case, a=4.5, b=-3, and c=12

plug in the values (± means that there are two solutions, one is when you add, and one is when you subtract.)

x=[-(-3)±√((-3)²-4*4.5*12)]/(2*4.5)

 =[3±√(9-216)]/9

 =[3±√(-207)]/9

√(-207)=√(207)*√(-1)=√(207)*i

√(207) can be simplified to 3√(23):

√(207)=√(9*23)=√(9)*√(23)=3√(23)

so now:

x=[3±3√(23)*i]/9

divide by 9 to get the answer in the correct form:

x=(3/9)±[(3√(23))/9]i

x=(1/3)±[√(23)/3]i

you can also rewrite this as x=(1/3)+[√(23)/3]i and x=(1/3)-[√(23)/3]i (just make sure to include both answers)

hopefully that helped! more practice will make these kinds of problems much easier :)