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Solve for [tex]$x$[/tex] in the equation [tex]$x^2 + 2x + 1 = 17$[/tex].

A. [tex][tex]$x = -1 \pm \sqrt{15}$[/tex][/tex]
B. [tex]$x = -1 \pm \sqrt{17}$[/tex]
C. [tex]$x = -2 \pm 2\sqrt{5}$[/tex]
D. [tex][tex]$x = -1 \pm \sqrt{13}$[/tex][/tex]



Answer :

GinnyAnswer
To solve the equation [tex]\(x^2 + 2x + 1 = 17\)[/tex], you can use the quadratic formula, which is [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]. Let's go through the steps to solve this quadratic equation.

1. Rewrite the equation in standard form [tex]\(ax^2 + bx + c = 0\)[/tex]:

[tex]\(x^2 + 2x + 1 = 17\)[/tex]

Subtract 17 from both sides to get:

[tex]\(x^2 + 2x + 1 - 17 = 0\)[/tex]

Simplifying, we get:

[tex]\(x^2 + 2x - 16 = 0\)[/tex]

2. Identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:

Here,
[tex]\(a = 1\)[/tex],
[tex]\(b = 2\)[/tex],
[tex]\(c = -16\)[/tex]

3. Calculate the discriminant [tex]\(\Delta = b^2 - 4ac\)[/tex]:

[tex]\(\Delta = 2^2 - 4(1)(-16) = 4 + 64 = 68\)[/tex]

4. Find the square root of the discriminant [tex]\(\sqrt{\Delta}\)[/tex]:

[tex]\(\sqrt{68} \approx 8.246211251235321\)[/tex]

5. Calculate the two solutions [tex]\(x_1\)[/tex] and [tex]\(x_2\)[/tex] using the quadratic formula:

[tex]\[ x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-2 + 8.246211251235321}{2 \cdot 1} = \frac{6.246211251235321}{2} \approx 3.1231056256176606 \][/tex]

[tex]\[ x_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-2 - 8.246211251235321}{2 \cdot 1} = \frac{-10.24621125123532}{2} \approx -5.123105625617661 \][/tex]

So, the solutions to the equation [tex]\(x^2 + 2x + 1 = 17\)[/tex] are approximately [tex]\(3.1231056256176606\)[/tex] and [tex]\(-5.123105625617661\)[/tex].

None of the provided choices match exactly these numerical solutions, indicating that the provided choices might contain errors. The correct solutions, based on our step-by-step calculation, do not match any of the given options.

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