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Chapter 10: Problem 42

Solve. $$x^{2}+4 x=60$$

Short Answer

Expert verified
x = -10 or x = 6

Step by step solution

01

Move constant term to the other side

To solve the quadratic equation, first move the constant term by subtracting 60 from both sides. So, the equation becomes: \[ x^2 + 4x - 60 = 0 \]
02

Factor the quadratic equation

Find two numbers that multiply to -60 and add to 4. These numbers are 10 and -6. Factor the equation: \[ (x + 10)(x - 6) = 0 \]
03

Solve for x

Set each factor equal to zero and solve for x: \[ x + 10 = 0 \] \[ x = -10 \] \[ x - 6 = 0 \] \[ x = 6 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

factoring quadratic equations
Factoring quadratic equations is a method used to break down the equation into simpler expressions. A quadratic equation is typically in the form of \(ax^2 + bx + c = 0\), where \a\, \b\, and \c\ are constants. In the equation provided, \(x^2 + 4x - 60 = 0\), \a = 1\, \b = 4\, and \c = -60\.

To factor this equation, we look for two numbers that multiply to \-60\ and add up to \4\. In this case, those numbers are \10\ and \-6\ because \10 \times (-6) = -60\ and \10 + (-6) = 4\.

Once these numbers are found, the equation can be written in its factored form as: \ (x + 10)(x - 6) = 0 \. Factoring helps to simplify the equation and make it easier to find the solutions.
setting equations to zero
Setting the equation to zero is a crucial step in solving quadratic equations. This involves moving all terms to one side of the equation to ensure the right side equals zero.

For the exercise provided, the original equation is \(x^2 + 4x = 60\). The constant term \60\ is moved to the left side by subtracting it from both sides, resulting in \(x^2 + 4x - 60 = 0\).

Having the equation set to zero is essential for the factoring method to work, as it allows us to find the points where the expression equals zero, leading to the solutions of the quadratic equation.

This process turns the equation into a format that is more straightforward to solve.
finding roots
Finding the roots of a quadratic equation means determining the values of \x\ that satisfy the equation. After factoring the quadratic equation and setting it to zero, the next step is to find the values of \x\ that make each factor zero.

In the provided example, we have: \ (x + 10)(x - 6) = 0 \. To find the roots, set each factor equal to zero:
- \(x + 10 = 0\), solving this gives \(x = -10\).
- \(x - 6 = 0\), solving this gives \(x = 6\).

Therefore, the roots of the quadratic equation \(x^2 + 4x - 60 = 0\) are \-10\ and \6\.

These values are also known as the solutions, zeros, or x-intercepts of the equation. Finding roots is the ultimate goal when solving quadratic equations, as it provides the values at which the quadratic expression equals zero.

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Most popular questions from this chapter

Factor completely. $$ 9 x^{2} y^{2}-30 x y+25 $$

Simplify. $$16^{-1 / 2}$$

The standard form for equations of horizontal or vertical hyperbolas centered at \((h, k)\) are as follows: $$ \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 $$ (Graph can't copy) $$ \frac{(y-k)^{2}}{b^{2}}-\frac{(x-h)^{2}}{a^{2}}=1 $$ The vertices are as labeled and the asymptotes are $$ y-k=\frac{b}{a}(x-h) \text { and } y-k=-\frac{b}{a}(x-h) $$ For each of the following equations of hyperbolas, complete the square, if necessary, and write in standard form. Find the center, the vertices, and the asymptotes. Then graph the hyperbola. $$ \frac{(x-5)^{2}}{36}-\frac{(y-2)^{2}}{25}=1 $$

How can you tell from the equation of an ellipse whether its graph is horizontal or vertical?

The equation \(x^{2}+y^{2}=\frac{81}{4},\) where \(x\) and \(y\) represent the number of meters from the center, can be used to draw the outer circle on a wrestling mat used in International, Olympic, and World Championship wrestling. The equation \(x^{2}+y^{2}=16\) can be used to draw the inner edge of the red zone. Find the area of the red zone. (IMAGE CANNOT COPY)
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