Question

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

Answer 1

Answer:

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

Step-by-step 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:

$\sqrt{(x+1)^2} =\sqrt{17}$

Now Square root and square gets cancelled.

$x+1=\sqrt{17} \\x=-1+\sqrt{17}$

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

$x=-1\±\sqrt{17}$

Hence  Final Answer is $x=-1\±\sqrt{17}$.

Answer 2

Answer:

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

Step-by-step explanation: