Let u=XY where X(-4, 6) and Y(8,3). What is 12 ul? 6/17 18 3/17 612
2 answers:
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1) x² - x - 12 = 0
Factor it
(x-4)(x+3) = 0
Study the factors
x-4 = 0 -> x =4
x+3 = 0 -> x = -3
2) t = x²
x⁴ = (x²)²
t² + 3t - 4 = 0
Use quadratic formula:
(-b±√b²-4ac)/2a
(-3±√9+16)/2
(-3±5)/2
t1 = 2/2 = 1
t2 = -8/2 = -4 (not possible, we choose t1)
So x² = 1
Result: x = <span>±1
3) Square both sides
2x-4 = (x-6)</span>²
2x-4 = x²-12x+36
Take all terms to the left
2x - 4 - x² + 12x - 36 = 0
-x² + 14x - 40 = 0
x² - 14x + 40 = 0
Use quadratic formula:
(-b±√b²-4ac)/2a
(14±√196-160)/2
(14±6)/2
x1 = 10
x2 = 4
Let's check
√2(10) - 4 = 10 - 6
√20-4 = 4
√16 = 4
4 = 4 (OK)
√2(4) - 4 = 4 - 6
√8-4 = -2
√4 = -2
2 = -2 (NO)
So x = 10
4) Factor x²-9 = (x-3)(x+3)
So the common denominator would be (x-3)(x+3)
Cancel out the denominators...
x+3 ≠ 0 -> x ≠ -3
x-3 ≠ 0 -> x ≠ 3
Take all members to the left
x(x+3)+x(x-3)-2 = 0
x²+3x+x²-3x-2 = 0
2x² - 2 = 0
2x² = 2
x² = 1
x = ±1
Factor it
(x-4)(x+3) = 0
Study the factors
x-4 = 0 -> x =4
x+3 = 0 -> x = -3
2) t = x²
x⁴ = (x²)²
t² + 3t - 4 = 0
Use quadratic formula:
(-b±√b²-4ac)/2a
(-3±√9+16)/2
(-3±5)/2
t1 = 2/2 = 1
t2 = -8/2 = -4 (not possible, we choose t1)
So x² = 1
Result: x = <span>±1
3) Square both sides
2x-4 = (x-6)</span>²
2x-4 = x²-12x+36
Take all terms to the left
2x - 4 - x² + 12x - 36 = 0
-x² + 14x - 40 = 0
x² - 14x + 40 = 0
Use quadratic formula:
(-b±√b²-4ac)/2a
(14±√196-160)/2
(14±6)/2
x1 = 10
x2 = 4
Let's check
√2(10) - 4 = 10 - 6
√20-4 = 4
√16 = 4
4 = 4 (OK)
√2(4) - 4 = 4 - 6
√8-4 = -2
√4 = -2
2 = -2 (NO)
So x = 10
4) Factor x²-9 = (x-3)(x+3)
So the common denominator would be (x-3)(x+3)
Cancel out the denominators...
x+3 ≠ 0 -> x ≠ -3
x-3 ≠ 0 -> x ≠ 3
Take all members to the left
x(x+3)+x(x-3)-2 = 0
x²+3x+x²-3x-2 = 0
2x² - 2 = 0
2x² = 2
x² = 1
x = ±1
