Find two numbers which differ by 4 and whose product is 45.
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calendarEducator since 2011
write2 answers
starTop subject is Math
Let's call the first number N1 and the second number N2.
So, let's call N1 the smaller number and N1=x
Let's call N2 the larger number and N2=x+4.
X(X+4)=45
x^2+4x=45
x^2+4x-45=0
(x+9)(x-5)=0
X=-9, X=5
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calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
We will assume that the numbers are x and y.
Given that the difference between the numbers is 4.
==> x - y = 4 .............(1)
Also, given the product of the numbers is 45.
==> x*y = 45................(2)
Now we will use the substitution method to solve.
From (1), we know that x = y + 4.
==> x*y = 45
==> (y+4)*y = 45
==> y^2 + 4y = 45
==> y^2 + 4y - 45 = 0
Now we will solve the quadratic equation using factors.
==> (y+ 9)(y-5) = 0
==> y1= -9 ==> x1= -5
==> y2 = 5 ==> x2= 9
Then the numbers are 5 and 9 OR -5 and -9.
calendarEducator since 2010
write12,551 answers
starTop subjects are Math, Science, and Business
The two numbers to be determined have a difference of 4, so let them be A and A + 4.
Now we are given that their product is 45
=> A*( A + 4) = 45
=> A^2 + 4A = 45
=> A^2 + 4A - 45 = 0
=> A^2 + 9A - 5A - 45 = 0
=> A( A + 9) - 5(A + 9) = 0
=> (A - 5)(A +9) = 0
This gives A = 5 and A = -9
So the two numbers are 5 , 9 and -9, -5
The required numbers can be (5, 9) and (-9, -5)
Let x and x+ 4 be the two numbers which differ by 4. Their product is 45.
So x(x+4) = 45.
=> x^2+4x-45 = 0
(x-5)(x+9)) = 0.
So x - 5 = 0, or x+9 = 0.
So x -5 = 0 gives x = 5. So the other number is x+4 = 9.
So x+ 9 = 0 gives x = -9 and the other number is -9+4 = -5.
Therefore the solution is (5 and 9) or ( -9 and -5).
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