Solve Quadratic Equation: X^2 - 14x + 31 = 63

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Understanding how to solve quadratic equations is a fundamental skill in mathematics, and today, we're going to tackle a specific one: $x^2 - 14x + 31 = 63$. Quadratic equations, characterized by their highest power of the variable being two (like $x^2$), often appear in various fields, from physics to engineering and economics. They can describe parabolas, projectile motion, and optimization problems. The general form of a quadratic equation is $ax^2 + bx + c = 0$. Our task is to find the value(s) of $x$ that make this equation true. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. The best method often depends on the specific equation. For our equation, $x^2 - 14x + 31 = 63$, the first step is to set it to the standard form $ax^2 + bx + c = 0$. To do this, we need to subtract 63 from both sides of the equation:

$x^2 - 14x + 31 - 63 = 63 - 63$

$x^2 - 14x - 32 = 0$

Now we have our equation in standard form, with $a=1$, $b=-14$, and $c=-32$. Let's explore a few methods to find the solutions for $x$.

Method 1: Factoring

Factoring is often the quickest method if it's possible. We are looking for two numbers that multiply to $c$ (which is -32 in our case) and add up to $b$ (which is -14). Let's list pairs of factors for -32 and check their sums:

• 1 and -32: Sum = -31
• -1 and 32: Sum = 31
• 2 and -16: Sum = -14
• -2 and 16: Sum = 14
• 4 and -8: Sum = -4
• -4 and 8: Sum = 4

We found a pair! The numbers 2 and -16 multiply to -32 and add up to -14. This means we can factor our quadratic equation as follows:

$(x + 2)(x - 16) = 0$

For this product to be zero, at least one of the factors must be zero. So, we set each factor equal to zero and solve for $x$:

$x + 2 = 0 x - 16 = 0$

$x = -2 x = 16$

So, the solutions are $x = -2$ and $x = 16$. This method was quite straightforward for this particular equation.

Method 2: Completing the Square

Completing the square is a powerful technique that can solve any quadratic equation, even if it's not easily factorable. The goal is to manipulate the equation into the form $(x-h)^2 = k$. We start with our equation in standard form: $x^2 - 14x - 32 = 0$. First, move the constant term to the right side:

$x^2 - 14x = 32$

Next, we need to add a specific value to both sides to make the left side a perfect square trinomial. This value is found by taking half of the coefficient of the $x$ term (which is -14), squaring it, and adding it to both sides. Half of -14 is -7, and $(-7)^2$ is 49.

$x^2 - 14x + 49 = 32 + 49$

The left side now factors as $(x - 7)^2$. The right side simplifies to 81.

$(x - 7)^2 = 81$

Now, we take the square root of both sides:

$\sqrt{(x - 7)^2} =

\pm

\sqrt{81}$

$x - 7 =

\pm

9$

Finally, we isolate $x$ by adding 7 to both sides:

$x = 7

\pm

9$

This gives us two possible solutions:

$x_1 = 7 + 9 = 16$

$x_2 = 7 - 9 = -2$

Once again, we arrive at the solutions $x = 16$ and $x = -2$. This method is very systematic and always works.

Method 3: The Quadratic Formula

The quadratic formula is a universal solution for any quadratic equation of the form $ax^2 + bx + c = 0$. The formula is:

$x =

\frac{{-b

\pm

\sqrt{{b^2 - 4ac}}}}{{2a}}$

In our equation, $x^2 - 14x - 32 = 0$, we have $a=1$, $b=-14$, and $c=-32$. Let's substitute these values into the formula:

$x =

\frac{{-(-14)

\pm

\sqrt{{(-14)^2 - 4(1)(-32)}}}}{{2(1)}}$

$x =

\frac{{14

\pm

\sqrt{{196 + 128}}}}{{2}}$

$x =

\frac{{14

\pm

\sqrt{{324}}}}{{2}}$

The square root of 324 is 18. So, the equation becomes:

$x =

\frac{{14

\pm

18}}{{2}}$

Now, we find the two possible values for $x$:

$x_1 =

\frac{{14 + 18}}{{2}} =

\frac{{32}}{{2}} = 16$

$x_2 =

\frac{{14 - 18}}{{2}} =

\frac{{-4}}{{2}} = -2$

All three methods yield the same solutions: $x = 16$ and $x = -2$. This consistency reinforces our confidence in the results.

Conclusion and Answer Choice Verification

We have successfully solved the quadratic equation $x^2 - 14x + 31 = 63$ using three different methods: factoring, completing the square, and the quadratic formula. Each method led us to the same pair of solutions: $x = 16$ and $x = -2$. Now, let's look at the provided answer choices:

A. $x=-16$ or $x=2$ B. $x=-7+3

\sqrt{7}$ C. $x=-2$ or $x=16$ D. $x=7

\pm

3

\sqrt{7}$

Our calculated solutions, $x=-2$ and $x=16$, directly match Option C. The other options represent different potential outcomes, likely stemming from common errors in calculation or formula application. For instance, option B and D involve square roots, indicating that the discriminant ($b^2 - 4ac$) was not a perfect square, which is not the case for our equation after simplification. Option A has the correct numbers but incorrect signs.

For further exploration into quadratic equations and their properties, you can visit resources like Khan Academy's Algebra section or MathWorld's page on Quadratic Equations.