For what value of x is the double product of the binomials x+2 and x–2 is 16 less than the sum of their squares?
Actually, a double product means two times.
So it would be
2(x+2)(x-2) = (x+2)^2 + (x-2)^2 - 16
...
The answer is AllRealNumbers
allrealnumbers
(x+2)(x-2) = (x+2)^2 + (x-2)^2 - 16
Not sure what a "double product" is, so if I got it wrong, just fix it and go on.
steve out here on every single problem like calm down
Well, if we multiply the binomials (x+2)(x-2), we get x^2 - 4. And if we add the squares of the binomials (x+2)^2 and (x-2)^2, we get x^2 + 4x + 4 + x^2 - 4x + 4. Now we just need to set up the equation, so we have x^2 - 4 = x^2 + 4x + 4 + x^2 - 4x + 4 - 16. Simplifying that mess, we get x^2 - 4 = 2x^2 + 4. Trust me, it's simplified! Now, let's isolate x: -4 = x^2 + 4 - 2x^2. Combining like terms, we get -4 = -x^2 + 4. Rearranging, we have x^2 - 4 = -4. And what do we do when we have a squared term and a number on the other side? Drum roll, please... we take the square root! So, taking the square root of both sides, we get x = √(-4), or in simplified form, x = a nice imaginary number. So, the value of x that satisfies the given condition is somewhere in the realm of unicorns and rainbows, my friend!
To find the value of x in this problem, we need to set up the equation based on the information given.
The double product of the binomials (x+2)(x-2) is calculated by multiplying the two binomials together:
(x+2)(x-2) = x^2 - 4
The sum of their squares is found by squaring each binomial and then adding them together:
(x+2)^2 + (x-2)^2 = x^2 + 4x + 4 + x^2 - 4x + 4
Simplifying this equation gives:
2x^2 + 8 = x^2 + 4x + 4 + x^2 - 4x + 4
Combining like terms:
2x^2 - x^2 + x^2 + 4x - 4x + 4 - 4 - 8 = 0
Simplifying further:
x^2 = 4
To solve for x, take the square root of both sides:
x = ±2
Therefore, the value of x that satisfies the given condition is x = 2 or x = -2.