SOLUTION: Please help me solve this equation! 1.) Find two numbers with a sum of 4 and a product of 13. answer should be [2+3i] and [2-3i]

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Question 122616: Please help me solve this equation!
1.) Find two numbers with a sum of 4 and a product of 13.
answer should be [2+3i] and [2-3i]

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Let one of the numbers be represented by x and the other one by y
.
You are told that the sum of the two numbers is 4. This can be written in equation form as:
.
x+%2B+y+=+4
.
You are also told that the product of the two numbers is 13. This can be written in equation
form as:
.
x%2Ay+=+13
.
So you now have two equations, each with the two unknowns and you need to solve them. One way
to get a solution is to use substitution. Solve one of the equations by finding one of the
unknowns in terms of the other and substitute this result into the other equation.
.
Let's begin by solving the product equation for y in terms of x. Divide both sides of the
product equation by x and you get:
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y+=+13%2Fx
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Then in the sum equation you can substitute 13%2Fx for y and you get:
.
x+%2B+13%2Fx+=+4
.
Get rid of the denominator by multiplying both sides (all terms) of this equation by x
to get:
.
x%5E2+%2B+13+=+4x
.
Get this into standard quadratic form by subtracting 4x from both sides to convert the
equation to:
.
x%5E2+-+4x+%2B+13+=+0
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Solve this by using the quadratic formula. Since this equation is of the standard form:
.
ax%5E2+%2B+bx+%2B+c+=+0
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by comparing this standard form term-by-term with with the equation for this problem you
can see that a = 1, b = -4, and c = 13. The quadratic formula says that the solution to
a quadratic equation of the standard form is given by:
.
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
.
So all that you have to do is to substitute your values for a, b, and c into this solution
equation and you have:
.
x+=+%28-%28-4%29+%2B-+sqrt%28%28-4%29%5E2-4%2A1%2A13+%29%29%2F%282%2A1%29+
.
This simplifies to:
.
x+=+%284+%2B-+sqrt%2816-52%29%29%2F2+=%284%2B-sqrt%28-36%29%29%2F2
.
But sqrt%28-36%29+=+sqrt%2836%2Ai%5E2%29+=+sqrt%2836%29%2Asqrt%28i%5E2%29+=+6%2Ai
.
and substituting this result into the answer simplifies it to:
.
x+=+%284%2B-sqrt%28-36%29%29%2F2+=+%284+%2B-+6i%29%2F2+=+2+%2B-+3i
.
So two possible answers for x are x+=+2%2B3i and x+=+2-3i
.
To find the corresponding values of y, return to the sum equation which said that:
.
x+%2B+y+=+4
.
Substitute x+=+2%2B3i into this equation and you get:
.
2%2B3i%2By+=+4
.
Solve for y by subtracting 2+%2B+3i from both sides and you get:
.
y+=+4+-+2+-+3i+=+2+-+3i
.
This tells you that if x+=+2%2B3i then y+=+2+-+3i.
.
Next, solve for the other possible value of x, namely x+=+2-3i. Again, start with
the equation:
.
x+%2B+y+=+4
.
Substitute x+=+2-3i into this equation and you get:
.
2-3i%2By+=+4
.
Solve for y by subtracting 2+-+3i from both sides and you get:
.
y+=+4+-+2+%2B+3i+=+2+%2B+3i
.
This tells you that if x+=+2-3i then y+=+2+%2B+3i.
.
Now examining the two sets of answers you see that if one of the numbers (x or y) is 2%2B3i
then the other number is 2-3i. So in either case, that is the answer. One number is
2%2B3i and the other number is 2-3i.
.
Hope this helps you to understand the problem and one way that it can be solved.
.