document.write( "Question 201614: How do you know if a quadratic equation will have one, two, or no solutions? How do you find a quadratic equation if you are only given the solution? Is it possible to have different quadratic equations with the same solution? Explain. Provide your classmate’s with one or two solutions with which they must create a quadratic equation \n" ); document.write( "

Algebra.Com's Answer #151815 by jsmallt9(3758) $\"\"$ $\"About$

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1. How to determine the number of solutions.
\n" ); document.write( "Let's look at the quadratic formula:
\n" ); document.write( " $\"x+=+%28-b+%2B-+sqrt%28b%5E2+-+4ac%29%29%2F2a\"$
\n" ); document.write( "The key is the expression in the square root: $\"b%5E2+-+4ac\"$ . In general there are three cases:

$\"b%5E2+-4ac\"$ is positive. The square root of a positive number is also some positive number. So in the numerator of the quadratic formula we will get two values: (-b + the square root) and (-b - the square root). So when $\"b%5E2+-+4ac+%3E+0\"$ we get two solutions.
$\"b%5E2+-4ac\"$ is zero. The square root of zero is zero. So in the numerator we get (-b + 0) and (-b - 0). But both of these are equal to -b! So when $\"b%5E2+-+4ac+=+0\"$ we only get one solution.
$\"b%5E2+-4ac\"$ is negative. And what is the square root of a negative number? What can we square and get a negative number as an answer? Answer: Nothing. You cannot square any Real number and get a negative. So when $\"b%5E2+-4ac+%3C+0\"$ there are no solutions.

\n" ); document.write( "2. Finding the equation from the solution(s)
\n" ); document.write( "One way to find solutions from the equation is to factor it. For example, solving
\n" ); document.write( " $\"x%5E2-5x%2B6=0\"$
\n" ); document.write( "we factor it:
\n" ); document.write( " $\"%28x-2%29%28x-3%29=0\"$
\n" ); document.write( "For a product to be zero one of the factors must be zero. In other \"words\":
\n" ); document.write( "x-2 = 0 or x-3 =0
\n" ); document.write( "Solving these we get:
\n" ); document.write( "x=2 or x=3
\n" ); document.write( "Now what you want is to be able to do this in reverse. Well all the steps above are reversible. Therefore, if we have two solutions: $\"x+=+x%5B1%5D\"$ and $\"x+=+x%5B2%5D\"$ , the equation is going to be:
\n" ); document.write( " $\"%28x+-+x%5B1%5D%29%28x+-+x%5B2%5D%29+=+0\"$
\n" ); document.write( "Some examples:
\n" ); document.write( "Solution: x = 1 or x = 6
\n" ); document.write( "Equation: (x-1)(x-6) = 0 which gives $\"x%5E2+-7x+%2B+6+=+0\"$
\n" ); document.write( "Solution: x = 10 or x = 0
\n" ); document.write( "Equation: (x-10)(x-0) = 0 or (x-10)(x) = 0 which gives $\"x%5E2+-10x+=+0\"$
\n" ); document.write( "Solution: x = -3 or x = 1.7
\n" ); document.write( "Equation: (x-(-3))(x-1.7) = 0 or (x+3)(x-1.7)=0 (You multiply out this one!?)
\n" ); document.write( "If you are given that there is only one solution to a quadratic equation then the equation is of the form: $\"%28x+-+x%5B1%5D%29%5E2+=+0\"$ . For exmaple, if the only solution to to a quadratic equation is 20, then the equation would be: $\"%28x+-+20%29%5E2+=+0\"$ which gives $\"x%5E2+-40x+%2B+400+=+0\"$ .

\n" ); document.write( "3. Can different quadratic equations have the same solution? Well it depends on what you mean by \"different\". Yes. For example: $\"%28x-4%29%28x-5%29+=+0\"$ $\"2%28x-4%29%28x-5%29+=+0\"$ and $\"-5%28x-4%29%28x-5%29+=+0\"$ all have same solution: x-4 or x=5. But are these \"different\" equations? They sure look different. But if you divide both sides of the second equation by 2 you get the first equation. If you divide both sides of the third equation by -5 you get the first equation. So are these equations \"different\"? In my opinion these equations are not different and that the answer to question #3 is no.\r
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