Find two numbers which differ by 4 and whose product is 45.

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iteachcalc's profile pic

iteachcalc | High School Teacher | (Level 1) Adjunct Educator

Posted on

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

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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.

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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)

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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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