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Find the real solutions of the following equation by graphing. x^3-9x^2+20x=0Asked by zaynab
Find the real solutions of the following equation by graphing.
x^2+8x^2+15x=0
x^2+8x^2+15x=0
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GPT 3.5
To graph the equation, we can first simplify it by combining like terms:
x^2 + 8x^2 + 15x = 0
9x^2 + 15x = 0
Next, we can factor out common terms:
x(9x + 15) = 0
Setting each factor equal to zero and solving for x, we find the possible values for x:
x = 0
9x + 15 = 0
9x = -15
x = -15/9
x = -5/3
Now we can plot these points on a graph.
The x-intercepts of the graph are x = 0 and x = -5/3.
Graphing the equation:
- Start by drawing a coordinate plane.
- Mark the x-intercepts at x = 0 and x = -5/3.
- Plot a curve that goes through these points.
The graph of the equation is a parabola that opens upwards. Since the equation is equal to zero, the parabola touches the x-axis at these two points.
Therefore, the real solutions of the equation are x = 0 and x = -5/3.
x^2 + 8x^2 + 15x = 0
9x^2 + 15x = 0
Next, we can factor out common terms:
x(9x + 15) = 0
Setting each factor equal to zero and solving for x, we find the possible values for x:
x = 0
9x + 15 = 0
9x = -15
x = -15/9
x = -5/3
Now we can plot these points on a graph.
The x-intercepts of the graph are x = 0 and x = -5/3.
Graphing the equation:
- Start by drawing a coordinate plane.
- Mark the x-intercepts at x = 0 and x = -5/3.
- Plot a curve that goes through these points.
The graph of the equation is a parabola that opens upwards. Since the equation is equal to zero, the parabola touches the x-axis at these two points.
Therefore, the real solutions of the equation are x = 0 and x = -5/3.
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