# The difference between the square of a negative number and five times the number is 40 more than the number. The number is? I think I set this up right, but how do I solve it ? (x)^2-5x=x+40...

The difference between the square of a negative number and five times the number is 40 more than the number. The number is? I think I set this up right, but how do I solve it ?

(x)^2-5x=x+40

Beginning and Intermediate Algebra, Martin-Gay 5th Ed. Chapter 11

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Let x be the unknown number.

Then `x^2-5x=40+x`

`x^2-6x-40=0`

`(x-10)(x+4)=0`

So by the zero product property either x=10 or x=-4.

Since we are asked for a negative number, x=-4.

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**The number is -4**

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It is given that the difference between the square of a negative number and five times the number is 40 more than the number.

Let the number be x. The square of the number is x^2. Five times the number is 5x.

As the difference between the square of a negative number and five times the number is 40 more than the number

x^2 - 5x = 40 + x

x^2 - 6x - 40 = 0

x^2 - 10x + 4x - 40 = 0

x(x - 10) + 4(x - 10) = 0

(x + 4)(x - 10) = 0

x = -4 and x = 10

As the number is negative take only the root x = -4.