Asked by Amanda
1. Explain in detail how to determine the value of the independent variable in a quadratic relation if the value of the dependent variable is known.
Please check my answer -
You would substitute the value for the dependent variable, then you would solve for the independent variable.
(is it right?)
2. What is the greatest number of solutions a quandratic equation can have? Explain, with an example, why all of the solutions to the equations may not be reasonable answers to the original problem.
My answer-
A quandratic equations can have either no solution, one solution or two solutions.
Can someone please help me with the answer to this question too? Please explain it also.. i find it so hard. >.<
Thanks!
Please check my answer -
You would substitute the value for the dependent variable, then you would solve for the independent variable.
(is it right?)
2. What is the greatest number of solutions a quandratic equation can have? Explain, with an example, why all of the solutions to the equations may not be reasonable answers to the original problem.
My answer-
A quandratic equations can have either no solution, one solution or two solutions.
Can someone please help me with the answer to this question too? Please explain it also.. i find it so hard. >.<
Thanks!
Answers
Answered by
drwls
1. The independent variable is the "x" in the quadratic equation ax^2 + bx + c = y (the dependent variable)
If you know y, then solve
ax^2 + bx + (c-y) = 0
c-y is a new constant, c'
The equation can be solved by completing the square or using the equation
x = [-b +/- sqrt(b^2-4ac')]/2a
2. You are right; the greatest number of solutions is two. There can be one solution in some cases. It is possible that all solutions contain imaginary numbers (the square root of a negative number), in which case there are no REAL solutions.
It is possible that only one of the two solutions is a reasonable solution to the problem. For example, if you throw a ball and ask when it hits the ground,
y = at - bt^2,
there will be two solutions; one will be the time t=0 when you threw it.
If you know y, then solve
ax^2 + bx + (c-y) = 0
c-y is a new constant, c'
The equation can be solved by completing the square or using the equation
x = [-b +/- sqrt(b^2-4ac')]/2a
2. You are right; the greatest number of solutions is two. There can be one solution in some cases. It is possible that all solutions contain imaginary numbers (the square root of a negative number), in which case there are no REAL solutions.
It is possible that only one of the two solutions is a reasonable solution to the problem. For example, if you throw a ball and ask when it hits the ground,
y = at - bt^2,
there will be two solutions; one will be the time t=0 when you threw it.
Answered by
Anonymous
Variable of 3/23=j/3
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