Asked by Lori
1.) Simplify the expression.
[sqrt3 + 2i][sqrt3 - 2i]
2.) Solve using the square root property. (y - 4)2 = -18
3.) Rewrite the the function in vertex form. You may use completing the square or an alternate method.
f(x) = 5x2 - 10x + 3
4.) Use the Discriminant b2 - 4ac to determine the type and number of solutions. 6= 4x - 5x2
Thanks! <3
[sqrt3 + 2i][sqrt3 - 2i]
2.) Solve using the square root property. (y - 4)2 = -18
3.) Rewrite the the function in vertex form. You may use completing the square or an alternate method.
f(x) = 5x2 - 10x + 3
4.) Use the Discriminant b2 - 4ac to determine the type and number of solutions. 6= 4x - 5x2
Thanks! <3
Answers
Answered by
Reiny
[sqrt3 + 2i][sqrt3 - 2i]
= 3 - 2i√3 + 2i√3 - 4i^2
= 3 - 4(-1) = 7
(y - 4)^2 = -18
y - 4 = ± √18
y = 4 ± 3√-2
= 4± 3√2 i
= 3 - 2i√3 + 2i√3 - 4i^2
= 3 - 4(-1) = 7
(y - 4)^2 = -18
y - 4 = ± √18
y = 4 ± 3√-2
= 4± 3√2 i
Answered by
Reiny
f(x) = 5x^2 - 10x + 3 , notice how we use exponents in this format
= 5(x^2 - 2x + 1 - 1) + 3
= 5 ( (x-1)^2 - 1) + 3
= 5(x-1)^2 - 5 - 3
= 5(x-1)^2 - 8
6= 4x - 5x^2
5x^2 - 4x + 6 = 0
b^2 - 4ac
= 16 - 4(5)(6) = -104
What do you think ?
= 5(x^2 - 2x + 1 - 1) + 3
= 5 ( (x-1)^2 - 1) + 3
= 5(x-1)^2 - 5 - 3
= 5(x-1)^2 - 8
6= 4x - 5x^2
5x^2 - 4x + 6 = 0
b^2 - 4ac
= 16 - 4(5)(6) = -104
What do you think ?
Answered by
Lori
For number 4 would the answer be "1 real"? Cause the options were 1 real, 2 real, or 2 complex.
Answered by
Reiny
No, since the disc I negative, you will have 2 complex roots
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