Square of a binomial x^2+2x+1

1 answer:

3 0

To put an equation into (x+c)^2, we need to see if the trinomial is a perfect square.
General form of a trinomial: ax^2+bx+c
If c is a perfect square, for example (1)^2=1, 2^2=4, that's a good indicator that it's a perfect square trinomial.
Here, it is, because 1 is a perfect square.
To ensure that it's a perfect square trinomial, let's look at b, which in this case is 2.
It has to be double what c is.
2 is the double of 1, therefore this is a perfect square trinomial.
Knowing this, we can easily put it into the form (x+c)^2.
And the answer is: (x+1)^2.
To do it the long way:
x^2+2x+1
Find 2 numbers that add to 2 and multiply to 1.
They are both 1.
x^2+x+x+1
x(x+1)+1(x+1)
Gather like terms
(x+1)(x+1)
or (x+1)^2.

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X² + y² = 8

ohaa [14]

Answer:

$(2,2)~ \text{and}~ (-2,-2)$

Step by step explanation:

$x^2 +y^2 =8~~~~~~...(i)\\\\x-y=0~~~~~~~~~...(ii)\\\\\text{From (ii):}\\\\x-y = 0 \implies x = y\\\\\text{Substitute x = y in eq (i):}\\\\~~~~~~y^2 +y^2 =8\\\\\implies 2y^2 = 8\\ \\\implies y^2 = \dfrac 82\\\\\implies y^2 = 4\\\\\implies y = \pm\sqrt 4\\\\\implies y = \pm 2\\\\\text{Substitute y = x in equation (i):}\\ \\~~~~~~x^2 +x^2 = 8\\ \\\implies 2x^2 = 8\\\\\implies x^2 = \dfrac 82\\\\\implies x^2 = 4\\\\\implies x = \pm\sqrt 4\\\\\implies x = \pm2\\\\$