Question

Solve for x in the equation x squared + 2 x + 1 = 17.

x = negative 1 plus-or-minus StartRoot 15 EndRoot
x = negative 1 plus-or-minus StartRoot 17 EndRoot
x = negative 2 plus-or-minus 2 StartRoot 5 EndRoot
x = negative 1 plus-or-minus StartRoot 13 EndRoot

Answer 1

B.

Explanation:

Got it right on Edge 2023. : )

Answer 2

x = negative 1 plus-or-minus StartRoot 17 EndRoot

Explanation:

Given:

`x^2+2x+1=17`

We need to solve for x.

`x^2+2x+1=17 \ \ \ \ equation\ 1`

First we will solve L.H.S

Factorizing the equation we get;

`x^2+2x+1\\=x^2+x+x+1\\=x(x+1)+1(x+1)\\=(x+1)(x+1)\\=(x+1)^2`

Now Things which are equal to the same thing are also equal to one another. Since

`x^2+2x+1=17`

`x^2+2x+1=(x+1)^2`

then, according to the law of transitivity,

`(x+1)^2=17`

Now, applying the Square Root Principle we get:

`√((x+1)^2) =√(17)`

Now Square root and square gets cancelled.

`x+1=√(17) \\x=-1+√(17)`

Since a square root has two values, one positive and the other negative.

`x=-1\±√(17)`

Hence Final Answer is
`x=-1\±√(17)`.