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Wa1337
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Homework Statement
P(x) = √6+5x-x2
Homework Equations
The Attempt at a Solution
P(x) = √6+5x-x2
-x2+5x+6 >/= 0
x2-5x-6 </= 0
(x-6)(x+1) </= 0
Wa1337 said:Homework Statement
P(x) = √6+5x-x2Homework Equations
The Attempt at a Solution
P(x) = √6+5x-x2
-x2+5x+6 >/= 0
x2-5x-6 </= 0
(x-6)(x+1) </= 0
To solve for x in this equation, you can use the quadratic formula or factor the equation to find the roots. You can also use a graphing calculator to visualize the solution points.
Solving for x in this equation helps us determine the values of x that make the equation true. This is important for finding the solutions to real-world problems and understanding the behavior of functions.
Yes, this equation can have up to two solutions for x. This is because it is a quadratic equation, which can have two real solutions, one real solution, or no real solutions depending on the discriminant.
The graph of this equation is a parabola that opens downwards and crosses the x-axis at the x-intercepts, which are the solutions for x in the equation.
Yes, since the equation contains a square root, the value inside the square root (6+5x-x2) must be greater than or equal to 0. This means that the values of x must fall within a certain range for the equation to have real solutions.