SOLUTION: I really need help with this problem. The problem says to find two consecutive intergers whose product is 90, it must be solved algebraically. Thank You.

Algebra ->  Real-numbers -> SOLUTION: I really need help with this problem. The problem says to find two consecutive intergers whose product is 90, it must be solved algebraically. Thank You.      Log On


   

Question 115555: I really need help with this problem. The problem says to find two consecutive intergers whose product is 90, it must be solved algebraically. Thank You.
Found 2 solutions by jim_thompson5910, bucky:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Remember, consecutive integers follow the pattern x, x%2B1, etc.

So if their product is 90, then

x%28x%2B1%29=90


x%5E2%2Bx=90 Distribute


x%5E2%2Bx-90=0 Move all of the terms to the left side


%28x%2B10%29%28x-9%29=0 Factor the left side (note: if you need help with factoring, check out this solver)



Now set each factor equal to zero:
x%2B10=0 or x-9=0

x=-10 or x=9 Now solve for x in each case


So our answer is
x=-10 or x=9




So our two numbers are either 9,10 or -10,-9

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Two unknown consecutive integers can be represented by x and x+1 since the second integer must
be one greater than the first one ... because they are consecutive.
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The product of these two integers is then given by:
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x*(x + 1)
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and this product must equal 90. This makes the equation become:
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x*(x+1) = 90
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And multiplying out the left side results in the equation:
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x^2 + x = 90
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Get rid of the 90 on the left side by subtracting 90 from both sides to get:
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x^2 + x - 90 = 0
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You can solve this by graphing it and seeing at what values of x the graph crosses the
x-axis. Or you can use the quadratic formula which is a short way of completing the square,
or you can factor it.
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Let's use the quadratic formula which says that for a quadratic equation of the standard form:
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ax^2 + bx + c = 0
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the values of x that satisfy this equation are given by:
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x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
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Compare our equation to the standard quadratic form. When you do you will see that "a" which
is the multiplier of x must be 1. b which is the multiplier of x must also be 1. And c which is
the constant must be -90. Substitute these values into the equation for x and you get:
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x+=+%28-1+%2B-+sqrt%28+1%5E2-4%2A1%2A-90+%29%29%2F%282%2A1%29+
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The term inside the radical sign multiplies out to become 1+%2B+360+=+361. This changes the
answer to:
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x+=+%28-1+%2B-+sqrt%28+361+%29%29%2F%282%2A1%29+
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but the square root of 361 is 19. So you can replace the radical by 19 and the answers for x
are then:
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x+=+%28-1+%2B-+19%29%2F%282%2A1%29+
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Notice that the denominator multiplies out to 2 which makes the problem:
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x+=+%28-1+%2B-+19%29%2F2+
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The two possible answers are then:
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x+=+%28-1-19%29%2F2+=+-20%2F2+=+-10 and
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x+=+%28-1+%2B+19%29%2F2+=+18%2F2+=+9
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If x is -10, then the next consecutive integer is -10 + 1 = -9. This means the consecutive
integers -10 and -9 are one answer.
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If x is +9, then the next consecutive integer is +9 + 1 = +10. This means the consecutive
integers +9 and +10 are another answer.
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Check: -10 times -9 = +90. That checks. And +9 times +10 = +90. That also checks. So
our two answers are correct.
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Hope this helps you with this problem.
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