# Find two consecutive even integers whose product is 288

## Answers

So I’m going to solve this using algebra OK?

So what you do is take the first integer as X.

Then we know that x(the first number) + X+2(the second number) is 288 right?

So we solve that saying x+x+2=288.

So 2X+2=288

2X+2-2=288-2 right?

So 2X=286.

So x is 286/2 right?

So x is 143.

Now the first number is 143 right?

So the second number should be 143+2 so 145.

I know they are not even but these are the solutions.

I don’t think there are any consecutive even numbers that add up to 288

(X)+(X+1)=24

"consecutive" means right after the other, so, if the first one is (X), the second integer must be (X+1). The answer 24 is given so you just have to solve the equation.

2X+1=24

2X=23

X=11.5

10 and 12 are two even integers whose product =120

Step-by-step explanation:

10 and 12 are two even integers whose product =120

[tex]10\times 12=120[/tex]

solve the answer

Step-by-step explanation:

let the two integers be x and x+1 as they are consecutive

so x(x+1) = 24

x^2 + x = 24

and now solve the quadratic equatioon

QUESTION 1

GIVEN:

[tex]long \: base = 3 \times (short \: base)[/tex]

[tex]height = (short \: base) - 2[/tex]

[tex]area = 30 {in}^{2}[/tex]

IMPORTANT EQUATIONS

[tex]trapezoid \: area = \frac{sum \: of \: bases}{2} \times height[/tex]

SOLVE:

[tex]30 = \frac{short + long}{2} (short - 2)[/tex]

[tex]30 = \frac{(short + 3short)}{2} \times (short - 2)[/tex]

[tex]30 = \frac{4short}{2} (short - 2)[/tex]

[tex]30 = 2{(short)}^{2} - 4(short)[/tex]

[tex]{short}^{2} - 2(short) - 15 = 0[/tex]

factor:

[tex](short - 5)(short + 3) = 0[/tex]

[tex]short = 3 \: and \: - 5[/tex]

Negative five is not a reasonable answer, so we focus on the positive three and say that is the length of the short base.

so, the answers:

[tex]short \: base = 3 \: in[/tex]

[tex]long \: base = 3 \times 3 = 9 \: in[/tex]

[tex]height = 3 - 2 = 1 \: in[/tex]

QUESTION 2

GIVEN

convert 3ft to inches.

[tex]l = 32 + w[/tex]

[tex]area = lw = 130 {in}^{2}[/tex]

or

[tex]w = \frac{130}{l}[/tex]

substitute

[tex]l = 32 + \frac{130}{l}[/tex]

solve for l:

[tex]{l}^{2} - 32 l- 130 = 0[/tex]

QUESTION 3

[tex]x(x + 2) = 80[/tex]

solve for X

[tex]{x}^{2} + 2x - 80 = 0[/tex]

[tex](x - 10)(x + 8) = 0[/tex]

[tex]x = 10 \: and \: - 8[/tex]

take the positive value since it's the only one that makes sense in this context.

so the two values are:

[tex]10 \: and \: 8[/tex]

You can't find the precise integer without knowing what the product is - your question is incomplete - but you can find a general solution.

If [tex]x[/tex] is one of the integers, then the next even integer is given by [tex]x+2[/tex]. So if the product is [tex]n[/tex], then you have

[tex]x(x+2)=k\implies x^2+2x=k\implies x^2+2x+1=k+1\implies (x+1)^2=k+1\implies x=-1\pm\sqrt{k+1}[/tex]

Plug in [tex]k[/tex] and select the solution for [tex]x[/tex] that is an even integer.