Which equation has the solutions x= 5 ± 2 square root 7 all over 3
1 answer:
Let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3
Then the equation is (x-x₁)(x-x₂) = 0
If you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.
You can see that a=1 (there is just an x²)
b = (-5 -2√7 + 2√7 - 5)/3 = -10/3
c = ((-5-2√7)/3 ) * (-5+2√7)/3) = -1/3
So the equation you're looking for is:
Then the equation is (x-x₁)(x-x₂) = 0
If you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.
You can see that a=1 (there is just an x²)
b = (-5 -2√7 + 2√7 - 5)/3 = -10/3
c = ((-5-2√7)/3 ) * (-5+2√7)/3) = -1/3
So the equation you're looking for is:
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The true statement about the function f(x) = -x² - 4x + 5 is that:
- The range of the function is all real numbers less than or equal to 9.
The domain of a function is the set of given values of input for which the function is valid and true.
The range is the dependent variable of a given set of values for which the function is defined.
- The domain of the function: f(x) = -x² - 4x + 5 are all real number from -∞ to +∞
For a parabola ax² + bx + c with the vertex
- If a < 0, then the range is f(x) ≤
- If a > 0, then the range f(x) ≥
- Here; a = -1,
The vertex for an up-down facing parabola for a function y = ax² + bx + c is:
Thus,
- vertex
= (-2, 9)
Range: f(x) ≤ 9
Therefore, we can conclude that the range of the function is all real numbers less than or equal to 9.
Learn more about the domain and range of a function here:
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