i've been stuck for over an hour and don't know what to do
i don't know is this right
? = "+" or "-"
# = number
y=x
(x+1)squared + (y+1)squared = 6
i substitute y
(x+1)squared + (x+1)squared = 6
x squared + 2x + 1 + x squared + 2x + 1 = 6
2x squared + 4x + 2 = 6
2x squared + 4x - 4 = 0
2(x squared + 2x - 2) = 0
2(x ? #)(x ? #)
i get to here and now i am stuck
can u correct me if i did something wrong
please and thank you
Reiny
answered
An easier way is not to expand it.
from
(x+1)^2 + (x+1)^2 = 6
2(x+)^2 = 6
(x+1)^2 = 3
x+1 = ± √3
x = -1 ± √3
take it from here
Explain Bot
answered
Starting with the equation:
(x+1)^2 + (x+1)^2 = 6
Expanding the squares:
x^2 + 2x + 1 + x^2 + 2x + 1 = 6
2x^2 + 4x + 2 = 6
Now, to simplify this equation and bring it to the general quadratic form (ax^2 + bx + c = 0), we need to move all the terms to one side and set it equal to zero.
2x^2 + 4x + 2 - 6 = 0
2x^2 + 4x - 4 = 0
So far, you've done everything correctly.
Next, let's factor out the common factor of 2:
2(x^2 + 2x - 2) = 0
Now, we need to factor the quadratic expression inside the parentheses to find the solutions for x.
There are different methods you can use to factor a quadratic expression, such as factoring by grouping, using the quadratic formula, or completing the square. In this case, we can try factoring by grouping.
The factors of -2 that add up to +2 are +4 and -2:
x^2 + 2x - 2 = (x + 4)(x - 2)
Substituting back into the equation, we have:
2(x + 4)(x - 2) = 0
Now, to find the values of x, we set each factor equal to zero and solve for x:
x + 4 = 0 --> x = -4
x - 2 = 0 --> x = 2
So, the solutions to the equation are x = -4 and x = 2.
I hope this helps!
