The product of 3 consecutive integers decreased by the cube of the first is 85. Wahat are the integers?
THE GIVEN EQUATION IS:
(x)(x+1)(x+2)-xcube = 85
How can i solve this ?
first step: expand the polynomial. You will see that x^3 goes away, leaving you with just
3x^2+2x = 85
Now just solve as you normally would:
3x^2 + 2x - 85 = 0
(3x+17)(x-5) = 0
To solve the equation (x)(x+1)(x+2) - x^3 = 85, you can follow these steps:
Step 1: Expand the expression (x)(x+1)(x+2) to get a quadratic equation.
(x)(x+1)(x+2) - x^3 = 85
(x^2+x)(x+2) - x^3 = 85
Step 2: Distribute and collect like terms.
x^3 + 2x^2 + x + 2x - x^3 = 85
2x^2 + 3x = 85
Step 3: Move all terms to one side to set the equation to zero.
2x^2 + 3x - 85 = 0
Step 4: Factor the quadratic equation. Since the leading coefficient is not 1, we can use the quadratic formula.
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 2, b = 3, and c = -85. Plugging these values into the formula, we have:
x = (-3 ± √(3^2 - 4(2)(-85))) / 2(2)
x = (-3 ± √(9 + 680)) / 4
x = (-3 ± √(689)) / 4
Step 5: Simplify the square root of 689, if possible.
The square root of 689 is an irrational number, so we leave it as √689.
Step 6: Solve for x by plugging the values of x back into the original equation.
Considering x = (-3 + √689) / 4:
First integer = x = (-3 + √689) / 4
Second integer = x + 1 = (-3 + √689 + 4) / 4
Third integer = x + 2 = (-3 + √689 + 8) / 4
Considering x = (-3 - √689) / 4:
First integer = x = (-3 - √689) / 4
Second integer = x + 1 = (-3 - √689 + 4) / 4
Third integer = x + 2 = (-3 - √689 + 8) / 4
Thus, there are two possible sets of consecutive integers:
1) (-3 + √689) / 4, (-3 + √689 + 4) / 4, and (-3 + √689 + 8) / 4
2) (-3 - √689) / 4, (-3 - √689 + 4) / 4, and (-3 - √689 + 8) / 4
To solve the equation (x)(x + 1)(x + 2) - x^3 = 85 and find the three consecutive integers, you can follow these steps:
Step 1: Expand the equation:
(x)(x + 1)(x + 2) - x^3 = 85
x(x + 1)(x + 2) - x^3 = 85
(x^2 + x)(x + 2) - x^3 = 85
x^3 + 2x^2 + x^2 + 2x - x^3 = 85
3x^2 + 2x = 85
Step 2: Simplify the equation:
3x^2 + 2x - 85 = 0
Step 3: Factorize or use the quadratic formula to solve for x. In this case, using the quadratic formula is more appropriate:
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 3, b = 2, and c = -85.
Plugging these values into the quadratic formula, we get:
x = (-2 ± √(2^2 - 4 * 3 * -85)) / (2 * 3)
Simplifying further:
x = (-2 ± √(4 + 1020)) / 6
x = (-2 ± √(1024)) / 6
x = (-2 ± 32) / 6
x = (30/6) or (-34/6)
Simplifying:
x = 5 or -17/3
Step 4: Find the consecutive integers:
Since we are looking for three consecutive integers, we can take the value of x and increment by 1 and 2 to find the other two integers.
For x = 5:
The three consecutive integers are 5, 6, and 7.
For x = -17/3:
The three consecutive integers are -17/3, -8/3, and -5/3.
So, the two sets of three consecutive integers that satisfy the given equation are 5, 6, 7 and -17/3, -8/3, -5/3.