1. squareroot of 25-squareroot of -16 divided by squareroot of 25+squareroot of -16
2. find the roots of x^3-5x^2-7x+51
2. find the roots of x^3-5x^2-7x+51
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1. Assume that the expression is (√25 - √(-16))/(√25 + √(-16))
Note that √25 = 5 and √-16 = 4i. This leaves us:
(5 - 4i)/(5 + 4i)
Multiply the top and bottom by the conjugate of 5 + 4i to get:
(5 - 4i)/(5 + 4i) * (5 - 4i)/(5 - 4i)
= (5 - 4i)²/(25 - 16i²)
= (25 - 40i + 16i²)/(25 + 16) [i² = -1]
= (25 - 16 - 40i)/41
= (9 - 40i)/41
2. Roots: -3 and 4 ± i. Use grouping method.
0 = x³ + 3x² - 8x² - 7x + 51
0 = x²(x + 3) - (8x² + 7x - 51)
0 = x²(x + 3) - (x + 3)(8x - 17)
0 = (x² - (8x - 17))(x + 3)
0 = (x + 3)(x² - 8x + 17)
Zero-product property:
x + 3 = 0 and x² - 8x + 17 = 0
x = {-3} and x² - 8x - 17 = 0
Quad. formula:
x = (8 ± √((-8)² - 4*1*17))/2
= (8 ± √(64 - 68))/2
= (8 ± √-4)/2
= (8 ± 2i)/2
= 4 ± i
I hope this helps!
Note that √25 = 5 and √-16 = 4i. This leaves us:
(5 - 4i)/(5 + 4i)
Multiply the top and bottom by the conjugate of 5 + 4i to get:
(5 - 4i)/(5 + 4i) * (5 - 4i)/(5 - 4i)
= (5 - 4i)²/(25 - 16i²)
= (25 - 40i + 16i²)/(25 + 16) [i² = -1]
= (25 - 16 - 40i)/41
= (9 - 40i)/41
2. Roots: -3 and 4 ± i. Use grouping method.
0 = x³ + 3x² - 8x² - 7x + 51
0 = x²(x + 3) - (8x² + 7x - 51)
0 = x²(x + 3) - (x + 3)(8x - 17)
0 = (x² - (8x - 17))(x + 3)
0 = (x + 3)(x² - 8x + 17)
Zero-product property:
x + 3 = 0 and x² - 8x + 17 = 0
x = {-3} and x² - 8x - 17 = 0
Quad. formula:
x = (8 ± √((-8)² - 4*1*17))/2
= (8 ± √(64 - 68))/2
= (8 ± √-4)/2
= (8 ± 2i)/2
= 4 ± i
I hope this helps!