Question

If 5 times a certain integer is subtracted from twice the square of the integer, the result is 63. Find the integer.​

Answer 1

Answer: 7 and -4.5

Explanation:

To solve this equation, we first need to put these terms into algebraic form. The first step would be to one side of the equation equal to 63. Then we are asked to subtract 5 times a number from another number. To do this we would put -5x. Then, we are given -5x is subtracted from twice an integer squared, so 2(
`x^(2)`) or 2
`x^(2)`.

Now, we can put all that together:

2
`x^(2)`-5x=63.

From here it's just solving the equation.

First subtract 63 on both sides so you get 2
`x^(2)`-5x-63=0

Then you factor the equation. Try substituting factors of 63 and 2 into the equation until you get an answer. The base of the factorization would be (2x+?)(x+?).

In this case, we find that 9 and -7 would work for ?'s.

(2x+9)(x-7)

when you multiply those together, you should get 2
`x^(2)`-5x=63.

Now we just solve both equations.

2x=-9, x=-4.5

x=7.

Now if we plug the answers back into the original equation 2
`(7)^(2)`-5(7)=63 or 2
`(-4.5)^(2)`-5(7)=63, we should get 63=63.

Therefore, the answer would be 7 and -4.5.

Hope that helped.

:)

Answer 2

x=7 or x=-9/2

Explanation:

the integer is x

2x^2-5x=63

2x^2-5x-63=0

2x^2 - 14x + 9x -63=0

2x(x-7) + 9(x-7)=0

(x-7)(2x+9)=0

x-7=0 or 2x+9=0

x=7 or x=-9/2