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Chapter 1: Problem 121

More on Solving Equations Find all real solutions of the equation. $$x^{2} \sqrt{x+3}=(x+3)^{3 / 2}$$

Short Answer

Expert verified
The solutions are \( x = \frac{1 + \sqrt{13}}{2} \) and \( x = \frac{1 - \sqrt{13}}{2} \).

Step by step solution

01

Simplify both sides

Notice that both sides of the equation have \( \sqrt{x+3} \). Rewrite the equation as \( x^2 \sqrt{x+3} = (x+3)^{3/2} = (x+3)\sqrt{x+3} \).
02

Cancel like terms

Divide both sides of the equation by \( \sqrt{x+3} \). This gives: \( x^2 = x + 3 \). Note that this division is valid only when \( x+3 eq 0 \), thus \( x eq -3 \).
03

Rearrange and solve the quadratic equation

Rearrange the equation as \( x^2 - x - 3 = 0 \); this is a standard quadratic equation. To solve it, use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] with \( a = 1, b = -1, c = -3 \).
04

Apply the quadratic formula

Calculate the discriminant \( b^2 - 4ac \): \( (-1)^2 - 4 \times 1 \times (-3) = 1 + 12 = 13 \). Now compute: \[ x = \frac{-(-1) \pm \sqrt{13}}{2 \times 1} = \frac{1 \pm \sqrt{13}}{2} \].
05

Verify the solutions

Since we divided by \( \sqrt{x+3} \) earlier, we must check that \( x = \frac{1 + \sqrt{13}}{2} \) and \( x = \frac{1 - \sqrt{13}}{2} \) satisfy the original issue. Remember \( x eq -3 \). Both of these values are greater than \(-3\), so they are valid solutions.

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Key Concepts

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

Quadratic Equations
Quadratic equations are second-order polynomial equations in a single variable with a general form of \( ax^2 + bx + c = 0 \). Here, \( x \) represents the unknown variable, while \( a \), \( b \), and \( c \) are known coefficients where \( a eq 0 \). If \( a = 0 \), the equation is linear, not quadratic.
  • The power of the variable is squared (hence the name quadratic).
  • These equations can have either two real roots, one real root, or two complex roots.
  • Solutions can often be found using factoring, completing the square, or the quadratic formula.
Quadratic equations appear frequently in various areas, from physics to finance, where they are used to calculate trajectories, optimize area, and more. They are fundamental in understanding more complex algebraic concepts.
Quadratic Formula
The quadratic formula provides a straightforward way to find the solutions of any quadratic equation. The formula is written as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This tool is especially helpful when a quadratic equation cannot be easily factored. Here's how it works:
  • The term \( -b \) indicates the opposite of the coefficient \( b \).
  • The symbol \( \pm \) means that there will be two solutions: one involves addition and the other subtraction.
  • The denominator \( 2a \) is twice the coefficient of the squared term.
The quadratic formula calculates the roots by evaluating the expression under the square root (the discriminant) and using it to determine real or complex roots.
Discriminant
The discriminant is a part of the quadratic formula that decides the nature of the roots of a quadratic equation. It is the expression under the square root sign in the quadratic formula: \( b^2 - 4ac \).Understanding the discriminant helps determine how many and what type of solutions exist:
  • If the discriminant is positive \((b^2 - 4ac > 0)\), there are two distinct real roots.
  • If the discriminant is zero \((b^2 - 4ac = 0)\), there is one real root (also known as a repeated or double root).
  • If the discriminant is negative \((b^2 - 4ac < 0)\), there are no real roots; the solutions are complex or imaginary numbers.
This component is vital in both theoretical and practical applications, where visualizing and predicting the nature of the roots is crucial.

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Most popular questions from this chapter

Radicals Simplify the expression, and eliminate any negative exponents(s). Assume that all letters denote positive numbers. (a) \(\sqrt[5]{x^{3} y^{2}} \sqrt[19]{x^{4} y^{16}}\) (b) \(\frac{\sqrt[3]{8 x^{2}}}{\sqrt{x}}\)

If \(a_{1}, a_{2}, \ldots, a_{n}\) are nonnegative numbers, then their arithmetic mean is \(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\) and their geometric mean is \(\sqrt[n]{a_{1} a_{2} \dots a_{n}}\). The arithmetic-geometric mean inequality states that the geometric mean is always less than or equal to the arithmetic mean. In this problem we prove this in the case of two numbers \(x\) and \(y .\) (a) If \(x\) and \(y\) are nonnegative and \(x \leq y,\) then \(x^{2} \leq y^{2}\) [ Hint: First use Rule 3 of Inequalities to show that \(\left.x^{2} \leq x y \text { and } x y \leq y^{2} .\right\\}\) (b) Prove the arithmetic-geometric mean inequality $$\sqrt{x y} \leq \frac{x+y}{2}$$

If you had a million \(\left(10^{6}\right)\) dollars in a suitcase, and you spent a thousand (10 ) dollars each day, how many years would it take you to use all the money? Spending at the same rate, how many years would it take you to empty a suitcase filled with a billion ( \(10^{9}\) ) dollars?

Radicals Simplify the expression, and eliminate any negative exponents(s). Assume that all letters denote positive numbers. (a) \(\sqrt[6]{y^{5}} \sqrt[3]{y^{2}}\) (b) \((5 \sqrt[3]{x})(2 \sqrt[4]{x})\)

If \(a
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