ABCD is a parallelogram. E is the point where the diagonals AC and BD meet. Prove that triangle ABE is congruent to triangle CDE
2 answers:
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Answer:
a) x₁ = 5 x₂ = -2
b) x ∈ ( -∞ , -2 ) ∪ ( 5 , +∞ )
c ) x ∈ ( - ∞ . -3 ) ∪ ( 4 . + ∞ )
d) x₁ = -5 x₂ = 2
e) x∈ [ -5 , 2 ]
f ) x = 0
g) x ∈ ( - ∞ , 0 )
Step-by-step explanation:
f(x) = x² - 3x -10
a) Solve f(x) = 0
x² - 3x -10 = 0 ⇒ factoring we look for two numbers ( a and b ) in such way that
a*b = - 10 and a + b = -3
The numbers are -5 and 2, then we can arrange the equation as follows
( x - 5 ) * ( x + 2 ) = 0
Solving that we get
x - 5 = 0 x = 5
x + 2 = 0 x = -2
b) f(x) > 0
( x - 5 ) * ( x + 2 ) See annex we graph roots of the equation and evaluate values of f(x) for points inside intervals, therefore
f(-3) = (-3 - 5 ) * ( -3 + 1 ) = (-8) * (-2) = 16 and conclude, all values smaller than -2 are valids solution for f(x) > 0
f(0) = (0 - 5) * ( 0 + 2 ) = -10
f(0) < 0 then there is not solution between -2 and 5
f(6) = ( 6 - 5 ) * ( 6 + 2 ) = 8
f(6) = 8 f(6) > 0
So we conclude
x ∈ ( -∞ , -2 ) ∪ ( 5 , +∞ )
c) f(x) ≥ 2
x² - 3x -10 ≥ 2
x² - 3x -10 -2 ≥ 0
x² - 3x -12 ≥ 0
Factoring we get
( x - 4 ) * ( x + 3 ) ≥ 0
f(-4) = -8 * -1 = 8 ≥ 0 all values small than -3 are solutions
f(0) = -4 * 3 = -12 between -3 and 4 there are not solutions
f(5) = 1 * 8 = 8 ≥ 0
So we conclude
x ∈ ( - ∞ . -3 ) ∪ ( 4 . + ∞ )
g(x) = x² + 3x - 10
g(x) = 0
x² + 3x - 10 = 0
Factoring
( x + 5 ) * ( x - 2 ) = 0
x₁ = -5
x₂ = 2
g(x) ≤ 0
( x + 5 ) * ( x - 2 ) ≤ 0
g(-6) = -1 * -8 = 8 ≥ 0 non solutions interval
g(0) = 5 * -2 = - 10 ≤ 0 solutions interval
g(3) = 8 * 1 = 8 ≥ 0 non solutions interval
We can see that the roots of the equation are valids solutions, then
x∈ [ -5 , 2 ]
f(x) = g(x)
x² - 3x -10 = x² + 3x - 10
- 6x = 0
x = 0
x² - 3x -10 > x² + 3x - 10
Evaluating for negative values
( -1)² - 3(-1) - 10 = 1 +3 - 10 = -6 f(-1)
( -1)² +3(-1) - 10 = 1 -3 -10 = -12 g(-1)
f(-1) > g(-1)
Evaluating for positive values
f(1) = (1)² - 3 (1) - 10 = - 12
g(1) = (1)² + 3 (1) - 10 = - 6
then g(1) > f(1)
f(0) = 10
g(0) = 10
Then f(x) > g(x) only for negative values or
f(x) > g(x) x ∈ ( - ∞ , 0 )
