Find 3 consecutive integers such that the square of the second increased by the third is 43.this is algebra.ty

Answer 1

Let the numbers be x-1, x, and x+1.

We can write the numbers as:
x^2 + (x+1) = 43 <== square of the second increased by the third

Solve for x:
x^2 + x - 42 = 0
(x + 7)(x - 6) = 0

We have two solutions for x: -7 and 6.

Test these numbers above.
x=6 means that your numbers are 5, 6, 7. 6^2 + 7 = 36 + 7 = 43 <==TRUE
x=-7 means that your numbers are -8, -7, -6. (-7)^2 + (-6) = 49 - 6 = 43 <==TRUE

You could use either 5,6,7 or -8,-7,-6. I suggest the positive numbers, unless your teacher wants all solutions.

Incidentally, you'll note I chose x-1, x, x+1 for the three consecutive integers. I did this on purpose because I knew I'd be squaring the second number. I could have easily gone with x, x+1, x+2, but then I would be working with (x+1)^2 + (x+2). While it's simple to work out, it's simpler with x^2 + (x+1). Sometimes, experience just allows you to see the best way to represent numbers.

Answer 2

Since you are squaring the second integer and adding the third integer to the second integer, call the second integer x.
Then your 3 consecutive integers are:
x-1, x, x+1.
Then: x^2 +(x+1)=43
x^2+x-42=0
To solve you can use the quadratic equation or, in this case, you can factor the equation to get:
(x+7)(x-6)=0 so x=-7,6
So the integers are 5,6,7 or -8,-7,-6.
You can multiply x^2+(x+1) to see that these integers work.
5,6,and7.

Answer 3

so, for the three consecutive integers, you can use

n+(n+1)+(n+2) ...see how n is the first, you add one to get the second...and same for the third

the square of the second would be (n+1)^2

increased by the third would give you (n+1)^2+(n+2)=43

to square the second,

(n+1)(n+1) FOIL first, inner, outer, last,

n^2+n+n+1=n^2+2n+1

add that to (n+2)

n^2+2n+1
+n +2

n^2+3n+3=43

subtract 43 from each side so you get 0 on the right

n^2+3n+3=43
-43 -43

n^2+3n-40=0

now factor the left side

(n+8)(n-5)= 0

since you have it factored and zero on the other side, you can

n+8=0 n-5=0
-8 -8 +5 +5

n=-8 n=5

so your integers are -8,-7, and -6

or

5, 6, 7

Answer 4

Take 3 consecutive integers as x , x+a million , x+2. (x+a million) + (x + 2) = 24 + (a million/2)x 2x+3 = 24 + (a million/2)x 2x+3 = (40 8 + x)/2 multiply the two sides by potential of two. 4x + 6 = 40 8 + x 3x = 40 two x = 14 the three consecutive integers are 14,15,sixteen Have a good day. :)

Answer 5

three integers are
x,
x+1,
x+2

(see they are consecutive)

now set up the formula.
(x+1)^2 + (x+2) = 43

simplify:
x ^2+ 2x +1 +x +2 =43
x^2 +3x - 40 = 0

now use the quadratic formula to solve for x.
or better yet, just take a guess and solve by trial and error.
actually if you take x=5, it works!
(25 + 15 - 40 = 0)
so the numbers are 5, 6, 7

Answer 6

the numbers are x-1,x and x+1
x^2 + x + 1 = 43
x^2 + x -42 = 0
(x - 6) (x + 7) = 0
so x = 6 or -7
so, 5 6 7

Answer 7

5

6*6=36+7=43

A=5,6,7