Solving Radical Equations Finding Values Of X

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Introduction

In the realm of mathematics, solving equations is a fundamental skill. Among various types of equations, radical equations, which involve square roots or other radicals, often pose a unique challenge. This article delves into the methods and techniques required to solve two such equations. We will explore the step-by-step process of isolating the radicals, squaring both sides, and verifying the solutions to ensure they are valid. This comprehensive guide aims to equip you with the knowledge and confidence to tackle radical equations effectively. Understanding how to solve radical equations is crucial not only for academic purposes but also for various applications in science, engineering, and other fields where mathematical modeling is essential.

(a) Solving the Equation

Initial Setup and Isolation of Radicals

To find the value of x that satisfies the equation 3x+1−2x−1=x+2,{ \sqrt{3x+1} - \sqrt{2x-1} = \sqrt{x+2}, } we first need to isolate one of the radicals. The given equation involves three square roots, making it a bit complex. A strategic approach is to move one of the square roots to the other side of the equation to simplify the squaring process. This isolation helps in reducing the complexity of the equation when we square both sides. By isolating a radical, we set the stage for eliminating the square root, which is a crucial step in solving radical equations. In this case, we can rewrite the equation as: 3x+1=2x−1+x+2.{ \sqrt{3x+1} = \sqrt{2x-1} + \sqrt{x+2}. } This rearrangement prepares us for the next step, where we will square both sides of the equation to eliminate one of the square roots. The goal here is to simplify the equation gradually, making it easier to solve for x. Remember, the key to solving radical equations is to eliminate the radicals through careful algebraic manipulation.

Squaring Both Sides: First Iteration

Now that we have isolated one radical, the next step is to square both sides of the equation. Squaring both sides is a fundamental technique in solving radical equations because it eliminates the square root, thus simplifying the equation. However, it's important to remember that squaring both sides can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original equation. Therefore, we must check our solutions at the end of the process. Squaring both sides of the equation 3x+1=2x−1+x+2{ \sqrt{3x+1} = \sqrt{2x-1} + \sqrt{x+2} } we get: (3x+1)2=(2x−1+x+2)2.{ (\sqrt{3x+1})^2 = (\sqrt{2x-1} + \sqrt{x+2})^2. } Expanding both sides, we have: 3x+1=(2x−1)+2(2x−1)(x+2)+(x+2).{ 3x+1 = (2x-1) + 2\sqrt{(2x-1)(x+2)} + (x+2). } This simplifies to: 3x+1=3x+1+2(2x−1)(x+2).{ 3x+1 = 3x+1 + 2\sqrt{(2x-1)(x+2)}. }

Further Simplification and Isolation

The equation has been simplified, but we still have a radical term. To proceed, we need to isolate the remaining square root term. This involves algebraic manipulation to get the square root term by itself on one side of the equation. By isolating the radical, we prepare for the next squaring operation, which will eliminate the remaining square root. This step is crucial for making the equation solvable. From the previous step, we have: 3x+1=3x+1+2(2x−1)(x+2).{ 3x+1 = 3x+1 + 2\sqrt{(2x-1)(x+2)}. } Subtracting 3x+1{3x+1} from both sides, we get: 0=2(2x−1)(x+2).{ 0 = 2\sqrt{(2x-1)(x+2)}. } Dividing both sides by 2, we further simplify to: 0=(2x−1)(x+2).{ 0 = \sqrt{(2x-1)(x+2)}. } This simplified equation is now ready for the next step, where we will square both sides again to eliminate the square root completely.

Squaring Both Sides: Second Iteration

We have now isolated the remaining radical, and the next step is to square both sides again. This will eliminate the square root, allowing us to solve for x. Squaring both sides is a critical step in solving radical equations, but it's important to remember that we must check our solutions at the end to ensure they are not extraneous. From the previous step, we have: 0=(2x−1)(x+2).{ 0 = \sqrt{(2x-1)(x+2)}. } Squaring both sides, we get: 02=((2x−1)(x+2))2,{ 0^2 = (\sqrt{(2x-1)(x+2)})^2, } which simplifies to: 0=(2x−1)(x+2).{ 0 = (2x-1)(x+2). } This gives us a quadratic equation that we can solve for x.

Solving the Quadratic Equation

Now we have a quadratic equation that we need to solve. Quadratic equations can be solved by factoring, completing the square, or using the quadratic formula. In this case, factoring is the most straightforward method. From the previous step, we have: 0=(2x−1)(x+2).{ 0 = (2x-1)(x+2). } To find the values of x that satisfy this equation, we set each factor equal to zero: 2x−1=0orx+2=0.{ 2x-1 = 0 \quad \text{or} \quad x+2 = 0. } Solving the first equation, 2x−1=0{2x-1 = 0}, we get: 2x=1,{ 2x = 1, } x=12.{ x = \frac{1}{2}. } Solving the second equation, x+2=0{x+2 = 0}, we get: x=−2.{ x = -2. } So, we have two potential solutions: x=12{x = \frac{1}{2}} and x=−2{x = -2}. However, we must check these solutions in the original equation to ensure they are not extraneous.

Verification of Solutions

After solving a radical equation, it's crucial to verify the solutions in the original equation. This is because squaring both sides can introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original one. To verify our solutions, we will substitute each potential solution back into the original equation and check if it holds true.

Our original equation is: 3x+1−2x−1=x+2.{ \sqrt{3x+1} - \sqrt{2x-1} = \sqrt{x+2}. } Let's check the first potential solution, x=12{x = \frac{1}{2}}: 3(12)+1−2(12)−1=12+2.{ \sqrt{3\left(\frac{1}{2}\right)+1} - \sqrt{2\left(\frac{1}{2}\right)-1} = \sqrt{\frac{1}{2}+2}. } Simplifying, we get: 32+1−1−1=12+2,{ \sqrt{\frac{3}{2}+1} - \sqrt{1-1} = \sqrt{\frac{1}{2}+2}, } 52−0=52,{ \sqrt{\frac{5}{2}} - \sqrt{0} = \sqrt{\frac{5}{2}}, } 52=52.{ \sqrt{\frac{5}{2}} = \sqrt{\frac{5}{2}}. } This solution is valid.

Now, let's check the second potential solution, x=−2{x = -2}: 3(−2)+1−2(−2)−1=−2+2.{ \sqrt{3(-2)+1} - \sqrt{2(-2)-1} = \sqrt{-2+2}. } Simplifying, we get: −6+1−−4−1=0,{ \sqrt{-6+1} - \sqrt{-4-1} = \sqrt{0}, } −5−−5=0.{ \sqrt{-5} - \sqrt{-5} = 0. } Since we have square roots of negative numbers, this solution is not valid in the real number system. Therefore, x=−2{x = -2} is an extraneous solution.

Final Solution for (a)

After solving the equation and verifying the solutions, we have found that only one solution is valid. The extraneous solution was introduced during the process of squaring both sides of the equation, which is a common occurrence when dealing with radical equations. Therefore, it is always necessary to check the solutions in the original equation to ensure their validity. In this case, we found that: x=12{ x = \frac{1}{2} } is the only valid solution for the equation: 3x+1−2x−1=x+2.{ \sqrt{3x+1} - \sqrt{2x-1} = \sqrt{x+2}. }

(b) Solving the Equation

Initial Setup and Isolation of the Radical

To find the value(s) of x that satisfy the equation 10x1−x=3,{ \sqrt{\frac{10x}{1-\sqrt{x}}} = 3, } we first need to isolate the radical expression. In this equation, the radical is already isolated on the left side, which simplifies our initial steps. The equation involves a square root containing a fraction, and within that fraction, there's another square root. This nested structure requires a careful approach to solve. The first step in solving radical equations is often to isolate the radical, which in this case is already done for us. This allows us to proceed directly to eliminating the square root by squaring both sides of the equation. Isolating the radical is a crucial step because it sets the stage for simplifying the equation and solving for the variable.

Squaring Both Sides

Since the radical is already isolated, the next step is to square both sides of the equation. Squaring both sides is a fundamental technique for eliminating square roots in equations. However, it's important to remember that this operation can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original equation. Therefore, we must check our solutions at the end of the process. Squaring both sides of the equation 10x1−x=3,{ \sqrt{\frac{10x}{1-\sqrt{x}}} = 3, } we get: (10x1−x)2=32.{ \left(\sqrt{\frac{10x}{1-\sqrt{x}}}\right)^2 = 3^2. } This simplifies to: 10x1−x=9.{ \frac{10x}{1-\sqrt{x}} = 9. } Now we have an equation without the main square root, but we still have a square root in the denominator. We need to address this to solve for x.

Eliminating the Denominator

To further simplify the equation, we need to eliminate the denominator. This involves multiplying both sides of the equation by the denominator, which will clear the fraction and make the equation easier to work with. Eliminating the denominator is a common algebraic technique used to simplify equations involving fractions. By doing so, we transform the equation into a more manageable form, allowing us to proceed with solving for the variable. From the previous step, we have: 10x1−x=9.{ \frac{10x}{1-\sqrt{x}} = 9. } Multiplying both sides by 1−x{1-\sqrt{x}}, we get: 10x=9(1−x).{ 10x = 9(1-\sqrt{x}). } This simplifies to: 10x=9−9x.{ 10x = 9 - 9\sqrt{x}. } Now we have an equation with a single square root term, which we need to isolate to proceed.

Isolating the Remaining Radical

We now need to isolate the remaining square root term. This involves rearranging the equation so that the term with the square root is on one side and all other terms are on the other side. Isolating the radical is a crucial step in solving radical equations because it prepares us for the next step, which is to square both sides again to eliminate the square root. By isolating the radical, we make the equation simpler and easier to solve. From the previous step, we have: 10x=9−9x.{ 10x = 9 - 9\sqrt{x}. } Adding 9x{9\sqrt{x}} to both sides and subtracting 10x{10x} from both sides, we get: 9x=9−10x.{ 9\sqrt{x} = 9 - 10x. } Now, we have the square root term isolated on one side of the equation. We are ready to proceed with squaring both sides again to eliminate the square root.

Squaring Both Sides Again

Now that we have isolated the remaining radical, we need to square both sides again to eliminate the square root. This step is crucial for solving the equation, but it's also important to remember that squaring both sides can introduce extraneous solutions. Therefore, we must check our solutions at the end to ensure they are valid. From the previous step, we have: 9x=9−10x.{ 9\sqrt{x} = 9 - 10x. } Squaring both sides, we get: (9x)2=(9−10x)2,{ (9\sqrt{x})^2 = (9 - 10x)^2, } which simplifies to: 81x=81−180x+100x2.{ 81x = 81 - 180x + 100x^2. } This gives us a quadratic equation that we can solve for x.

Rearranging into Quadratic Form

We now have an equation that is quadratic in nature. To solve it, we need to rearrange the equation into the standard quadratic form, which is ax2+bx+c=0{ax^2 + bx + c = 0}. This form allows us to easily apply methods such as factoring, completing the square, or using the quadratic formula to find the solutions. Rearranging the equation into quadratic form is a standard technique in algebra, and it is essential for solving quadratic equations. From the previous step, we have: 81x=81−180x+100x2.{ 81x = 81 - 180x + 100x^2. } Subtracting 81x{81x} from both sides, we get: 0=100x2−261x+81.{ 0 = 100x^2 - 261x + 81. } This is a quadratic equation in the standard form, where a=100{a = 100}, b=−261{b = -261}, and c=81{c = 81}. We can now solve this equation for x.

Solving the Quadratic Equation

Now we have a quadratic equation in the standard form, and we need to solve it for x. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, factoring might be challenging due to the large coefficients, so we will use the quadratic formula. The quadratic formula is a general solution for quadratic equations of the form ax2+bx+c=0{ax^2 + bx + c = 0}, and it is given by: x=−b±b2−4ac2a.{ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. } For our equation, 100x2−261x+81=0{100x^2 - 261x + 81 = 0}, we have a=100{a = 100}, b=−261{b = -261}, and c=81{c = 81}. Plugging these values into the quadratic formula, we get: x=−(−261)±(−261)2−4(100)(81)2(100).{ x = \frac{-(-261) \pm \sqrt{(-261)^2 - 4(100)(81)}}{2(100)}. } Simplifying, we have: x=261±68121−32400200,{ x = \frac{261 \pm \sqrt{68121 - 32400}}{200}, } x=261±35721200,{ x = \frac{261 \pm \sqrt{35721}}{200}, } x=261±189200.{ x = \frac{261 \pm 189}{200}. } This gives us two potential solutions: x1=261+189200=450200=94,{ x_1 = \frac{261 + 189}{200} = \frac{450}{200} = \frac{9}{4}, } x2=261−189200=72200=925.{ x_2 = \frac{261 - 189}{200} = \frac{72}{200} = \frac{9}{25}. } So, we have two potential solutions: x=94{x = \frac{9}{4}} and x=925{x = \frac{9}{25}}. However, we must check these solutions in the original equation to ensure they are not extraneous.

Verification of Solutions

After solving the quadratic equation, it's crucial to verify the solutions in the original equation. This is because squaring both sides can introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original one. To verify our solutions, we will substitute each potential solution back into the original equation and check if it holds true.

Our original equation is: 10x1−x=3.{ \sqrt{\frac{10x}{1-\sqrt{x}}} = 3. } Let's check the first potential solution, x=94{x = \frac{9}{4}}: 10(94)1−94=3.{ \sqrt{\frac{10\left(\frac{9}{4}\right)}{1-\sqrt{\frac{9}{4}}}} = 3. } Simplifying, we get: 9041−32=3,{ \sqrt{\frac{\frac{90}{4}}{1-\frac{3}{2}}} = 3, } 452−12=3,{ \sqrt{\frac{\frac{45}{2}}{-\frac{1}{2}}} = 3, } −45=3.{ \sqrt{-45} = 3. } Since we have a square root of a negative number, this solution is not valid in the real number system. Therefore, x=94{x = \frac{9}{4}} is an extraneous solution.

Now, let's check the second potential solution, x=925{x = \frac{9}{25}}: 10(925)1−925=3.{ \sqrt{\frac{10\left(\frac{9}{25}\right)}{1-\sqrt{\frac{9}{25}}}} = 3. } Simplifying, we get: 90251−35=3,{ \sqrt{\frac{\frac{90}{25}}{1-\frac{3}{5}}} = 3, } 18525=3,{ \sqrt{\frac{\frac{18}{5}}{\frac{2}{5}}} = 3, } 182=3,{ \sqrt{\frac{18}{2}} = 3, } 9=3,{ \sqrt{9} = 3, } 3=3.{ 3 = 3. } This solution is valid.

Final Solution for (b)

After solving the equation and verifying the solutions, we have found that only one solution is valid. The extraneous solution was introduced during the process of squaring both sides of the equation, which is a common occurrence when dealing with radical equations. Therefore, it is always necessary to check the solutions in the original equation to ensure their validity. In this case, we found that: x=925{ x = \frac{9}{25} } is the only valid solution for the equation: 10x1−x=3.{ \sqrt{\frac{10x}{1-\sqrt{x}}} = 3. }

Conclusion

Solving radical equations requires a systematic approach. We must isolate the radicals, square both sides, and, most importantly, verify the solutions to eliminate extraneous ones. The examples discussed illustrate the step-by-step process and the importance of checking solutions. Mastering these techniques is crucial for success in algebra and beyond. The process involves careful algebraic manipulation and a thorough understanding of the properties of radicals and equations. By following these steps and practicing regularly, you can develop the skills necessary to solve a wide range of radical equations effectively. Remember, the key is to be patient, methodical, and always verify your solutions.