Problem 112 $$ 3 \cdot \sqrt[x]{81}-10 \cd... [FREE SOLUTION]

Chapter 7: Problem 112

$$ 3 \cdot \sqrt[x]{81}-10 \cdot \sqrt[x]{9}+3=0 $$

Short Answer

Expert verified

We are given the equation $3 \cdot \sqrt[x]{81}-10 \cdot \sqrt[x]{9}+3=0$. By substituting $y = \sqrt[x]{3}$, we transform the given equation into a quadratic equation $3(y^2)^2 - 10y^2 + 3 = 0$. Using the quadratic formula, we find two solutions for $y^2$: $y^2_1 = 3$ and $y^2_2 = \frac{1}{3}$. We then find the values of $y$ and use the substitution relationship ($y^x = 3$) to solve for $x$. The two valid real solutions are $x = 2$ and $x = -2$.

Step by step solution

Substitution

Let $ y = \sqrt[x]{3} $, which means $y^x = 3$. Substitute this new variable into the given equation: $ 3 \cdot \sqrt[x]{81} - 10 \cdot \sqrt[x]{9} + 3= 0 $ becomes $ 3 \cdot (y^4) - 10 \cdot (y^2) + 3 = 0 $

Quadratic Equation

Now we can see that this equation is a quadratic equation in terms of $y^2$: $ 3(y^2)^2 - 10y^2 + 3 = 0 $

Solve the Quadratic Equation

To solve this quadratic equation, we can use the quadratic formula: $ y^2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ Where a = 3, b = -10, and c = 3. Calculating the discriminant: $ \Delta = b^2 - 4ac = (-10)^2 - 4 \cdot 3 \cdot 3 = 64 $ Now, find the two solutions for $y^2$: $ y^2_1 = \frac{-(-10) + \sqrt{64}}{2 \cdot 3} = \frac{10 + 8}{6} = 3 $ $ y^2_2 = \frac{-(-10) - \sqrt{64}}{2 \cdot 3} = \frac{10 - 8}{6} = \frac{1}{3} $

Find the Solutions for $y$

Since we found the values for $y^2$, we now need to find the corresponding values for $y$: For $y^2_1 = 3$: $ y_1 = \pm\sqrt{3} $ For $y^2_2 = \frac{1}{3}$: $y_2 = \pm\frac{1}{\sqrt{3}} $

Find the Solutions for $x$

Recall the substitution: $ y = \sqrt[x]{3} $ or $ y^x = 3 $. We'll use this to find the values of $x$: (1) For $y_1 = \sqrt{3}$: $ (\sqrt{3})^x = 3 $ or $ 3^{\frac{x}{2}} = 3 $. Comparing the powers, we get: $ x = 2 $ (2) For $y_1 = -\sqrt{3}$, the equation doesn't make sense for real numbers, so we skip it. (3) For $y_2 = \frac{1}{\sqrt{3}}$: $ (\frac{1}{\sqrt{3}})^x = 3 $ or $ 3^{-\frac{x}{2}} = 3 $. Comparing the powers, we get: $ -\frac{x}{2} = 1 $ or $ x = -2 $ (4) For $y_2 = -\frac{1}{\sqrt{3}}$, the equation doesn't make sense for real numbers, so we skip it.

State the Final Answer

We found two valid real solutions for $x$: $x = 2$ and $x = -2$.

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Key Concepts

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

Substitution Method

The substitution method is a powerful technique to simplify complex equations, such as those involving variables within roots or exponents. In this exercise, we began by letting $ y = \sqrt[x]{3} $ to transform a more complicated expression into a manageable quadratic equation.
By setting this substitution, we can rewrite the original equation as $ 3(y^4) - 10(y^2) + 3 = 0 $.
This allowed us to focus on solving for $ y^2 $, treating the problem as a familiar quadratic equation instead of dealing with more cumbersome expressions. This step is crucial for reducing complexity and making further calculations straightforward.

Quadratic Formula

The quadratic formula is a staple tool in solving quadratic equations of the form $ ax^2 + bx + c = 0 $.
Given coefficients $ a = 3 $, $ b = -10 $, and $ c = 3 $, the formula\[y^2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] provides us a systematic way to find the equation's roots based on these parameters. Substituting the values, we unfold the possible solutions for $ y^2 $, which in this exercise are $ 3 $ and $ \frac{1}{3} $. This formula transforms the challenge of solving quadratics into a direct computation, allowing us to handle even non-trivial problems efficiently.

Discriminant Calculation

The discriminant, expressed as $ \Delta = b^2 - 4ac $, is integral in determining the nature of the roots of a quadratic equation. In this problem, after substitution, we calculated $ \Delta $ to be 64. This positive value signifies that our quadratic equation has two distinct real roots.
In practice, the discriminant informs:

Two distinct real roots if $ \Delta > 0 $.
One real root if $ \Delta = 0 $.
No real roots (two complex roots) if $ \Delta < 0 $.

Thus, knowing $ \Delta $ enables predictions about the types of solutions we should expect, guiding our solution strategy effectively.

Roots of Quadratic Equation

Finding the roots of a quadratic equation reveals the solutions to the problem. Once $ y^2 = 3 $ and $ y^2 = \frac{1}{3} $ were determined, we derived the potential values for $ y $ by square rooting each term:

$ y_1 = \pm \sqrt{3} $
$ y_2 = \pm \frac{1}{\sqrt{3}} $

These outcomes offer four total cases, though, in practical terms, some solutions, like negative roots, may be invalid depending on the context. By equating $ \sqrt[x]{3} $, the exercise provided the final real roots for $ x $, which are $ x = 2 $ and $ x = -2 $. Determining these roots involves both algebraic manipulation and reasoning about the underlying mathematics of the substitution and original problem statement.

91影视

$$ 3 \cdot \sqrt[x]{81}-10 \cdot \sqrt[x]{9}+3=0 $$

Short Answer

Step by step solution

Substitution

Quadratic Equation

Solve the Quadratic Equation

Find the Solutions for \(y\)

Find the Solutions for \(x\)

State the Final Answer

Key Concepts

Substitution Method

Quadratic Formula

Discriminant Calculation

Roots of Quadratic Equation

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