Given That 3 X + 1 − 5 X − 1 = 4 3 − 3 X \sqrt{3x+1} - \sqrt{5x-1} = \frac{4}{3} - 3x 3 X + 1 ​ − 5 X − 1 ​ = 3 4 ​ − 3 X , Complete The Squares To Find The Value For Which The Equation Is Satisfied.

Introduction

In this article, we will explore the process of completing the squares to solve a given equation involving square roots. The equation is $\sqrt{3x+1} - \sqrt{5x-1} = \frac{4}{3} - 3x$. We will use the method of completing the squares to find the value of $x$ that satisfies the equation.

Step 1: Isolate the Square Root Terms

The first step is to isolate the square root terms on one side of the equation. We can do this by moving the $-3x$ term to the left-hand side of the equation.

$\sqrt{3x+1} - \sqrt{5x-1} + 3x = \frac{4}{3}$

Step 2: Move the Terms to the Left-Hand Side

Next, we will move the $\sqrt{5x-1}$ term to the left-hand side of the equation by subtracting it from both sides.

$\sqrt{3x+1} = \sqrt{5x-1} - 3x + \frac{4}{3}$

Step 3: Square Both Sides

Now, we will square both sides of the equation to eliminate the square roots.

$(\sqrt{3x+1})^2 = (\sqrt{5x-1} - 3x + \frac{4}{3})^2$

Step 4: Expand the Right-Hand Side

Next, we will expand the right-hand side of the equation using the formula $(a-b)^2 = a^2 - 2ab + b^2$.

$3x+1 = (5x-1) - 2(3x)(\sqrt{5x-1}) + (3x)^2 - 2(3x)(\frac{4}{3}) + (\frac{4}{3})^2$

Step 5: Simplify the Right-Hand Side

Now, we will simplify the right-hand side of the equation by combining like terms.

$3x+1 = 5x-1 - 6x\sqrt{5x-1} + 9x^2 - 8x + \frac{16}{9}$

Step 6: Move the Terms to the Left-Hand Side

Next, we will move the terms to the left-hand side of the equation by subtracting $5x-1$ from both sides.

$-2x+2 = -6x\sqrt{5x-1} + 9x^2 - 8x + \frac{16}{9}$

Step 7: Move the Terms to the Left-Hand Side

Next, we will move the terms to the left-hand side of the equation by subtracting $9x^2 - 8x$ from both sides.

$-2x+2 + 9x^2 - 8x = -6x\sqrt{5x-1}$

Step 8: Factor Out the Common Term

Next, we will factor out the common term $-2x$ from the left-hand side of the equation.

$-2x(1+4x-4) = -6x\sqrt{5x-1}$

Step 9: Simplify the Left-Hand Side

Now, we will simplify the left-hand side of the equation by combining like terms.

$-2x(4x-3) = -6x\sqrt{5x-1}$

Step 10: Divide Both Sides by $-2x$

Next, we will divide both sides of the equation by $-2x$ to eliminate the $-2x$ term.

$(4x-3) = \frac{-6x\sqrt{5x-1}}{-2x}$

Step 11: Simplify the Right-Hand Side

Now, we will simplify the right-hand side of the equation by canceling out the $-2x$ term.

$(4x-3) = 3\sqrt{5x-1}$

Step 12: Square Both Sides

Next, we will square both sides of the equation to eliminate the square root.

$(4x-3)^2 = (3\sqrt{5x-1})^2$

Step 13: Expand the Left-Hand Side

Next, we will expand the left-hand side of the equation using the formula $(a-b)^2 = a^2 - 2ab + b^2$.

$(4x-3)^2 = 16x^2 - 24x + 9$

Step 14: Expand the Right-Hand Side

Next, we will expand the right-hand side of the equation using the formula $(a-b)^2 = a^2 - 2ab + b^2$.

$(3\sqrt{5x-1})^2 = 9(5x-1)$

Step 15: Simplify the Right-Hand Side

Now, we will simplify the right-hand side of the equation by multiplying the terms.

$9(5x-1) = 45x - 9$

Step 16: Equate the Left-Hand Side and the Right-Hand Side

Next, we will equate the left-hand side and the right-hand side of the equation.

$16x^2 - 24x + 9 = 45x - 9$

Step 17: Move the Terms to the Left-Hand Side

Next, we will move the terms to the left-hand side of the equation by subtracting $45x$ from both sides.

$16x^2 - 69x + 18 = 0$

Step 18: Solve the Quadratic Equation

Now, we will solve the quadratic equation using the quadratic formula.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Step 19: Plug in the Values

Next, we will plug in the values $a = 16$, $b = -69$, and $c = 18$ into the quadratic formula.

$x = \frac{-(-69) \pm \sqrt{(-69)^2 - 4(16)(18)}}{2(16)}$

Step 20: Simplify the Expression

Now, we will simplify the expression inside the square root.

$x = \frac{69 \pm \sqrt{4761 - 1152}}{32}$

Step 21: Simplify the Expression

Now, we will simplify the expression inside the square root.

$x = \frac{69 \pm \sqrt{3609}}{32}$

Step 22: Simplify the Expression

Now, we will simplify the expression inside the square root.

$x = \frac{69 \pm 189}{32}$

Step 23: Solve for $x$

Now, we will solve for $x$ by plugging in the values $x = \frac{69 + 189}{32}$ and $x = \frac{69 - 189}{32}$.

$x = \frac{258}{32}$ or $x = \frac{-120}{32}$

Step 24: Simplify the Expression

Now, we will simplify the expression by dividing both the numerator and the denominator by their greatest common divisor.

$x = \frac{129}{16}$ or $x = \frac{-15}{4}$

Conclusion

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Q: What is the method of completing the squares?

A: The method of completing the squares is a technique used to solve equations involving square roots. It involves manipulating the equation to create a perfect square trinomial, which can be factored into a binomial squared.

Q: How do I know when to use the method of completing the squares?

A: You should use the method of completing the squares when you have an equation involving square roots and you want to eliminate the square roots. This method is particularly useful when you have an equation with two square root terms and you want to combine them into a single square root term.

Q: What are the steps involved in completing the squares?

A: The steps involved in completing the squares are:

1. Isolate the square root terms on one side of the equation.
2. Move the terms to the left-hand side of the equation.
3. Square both sides of the equation.
4. Expand the right-hand side of the equation.
5. Simplify the right-hand side of the equation.
6. Move the terms to the left-hand side of the equation.
7. Factor out the common term.
8. Simplify the left-hand side of the equation.
9. Divide both sides by the common term.
10. Square both sides of the equation again.
11. Expand the left-hand side of the equation.
12. Equate the left-hand side and the right-hand side of the equation.
13. Move the terms to the left-hand side of the equation.
14. Solve the quadratic equation.

Q: What is the quadratic formula?

A: The quadratic formula is a formula used to solve quadratic equations of the form $ax^2 + bx + c = 0$. The formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Q: How do I use the quadratic formula to solve a quadratic equation?

A: To use the quadratic formula to solve a quadratic equation, you need to plug in the values of $a$, $b$, and $c$ into the formula. You can then simplify the expression inside the square root and solve for $x$.

Q: What are the advantages of using the method of completing the squares?

A: The advantages of using the method of completing the squares are:

• It allows you to eliminate the square roots in an equation.
• It can be used to solve equations involving two square root terms.
• It can be used to solve quadratic equations.

Q: What are the disadvantages of using the method of completing the squares?

A: The disadvantages of using the method of completing the squares are:

• It can be a complex and time-consuming process.
• It requires a good understanding of algebra and quadratic equations.
• It may not be suitable for all types of equations.

Q: Can I use the method of completing the squares to solve equations involving other types of radicals?

A: No, the method of completing the squares is specifically designed to solve equations involving square roots. It may not be suitable for solving equations involving other types of radicals, such as cube roots or fourth roots.

Q: Can I use the method of completing the squares to solve equations involving complex numbers?

A: No, the method of completing the squares is specifically designed to solve equations involving real numbers. It may not be suitable for solving equations involving complex numbers.

Conclusion

In this article, we have answered some of the most frequently asked questions about the method of completing the squares. We have discussed the steps involved in completing the squares, the advantages and disadvantages of using this method, and the types of equations that can be solved using this method. We hope that this article has been helpful in providing you with a better understanding of the method of completing the squares.