Lesson HOW TO solve equations containing radicals
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<H2>How to solve equations containing radicals</H2> An equation is called a <B>radical equation</B> if it contains the variable under a square root, cube root etc. There are two major methods for solving equations containing radicals: 1) To isolate radical on one side of the equation and then to eliminate it by raising both sides to a power equal to the radical index, and 2) To introduce a new variable. Examples below show how to use these methods. <H3>Problem 1</H3>Solve an equation {{{root(3,2x-3) = 3}}}. <B>Solution</B> <pre> Raise each side to a 3-rd power (the index of the radical is 3). You get an equation 2x-3 = 27, 2x = 30. It has a solution x=15. You should check this solution. To check it, simply substitute the value x=15 into the original equation. You get {{{root(3,2*15-3)}}} = {{{root(3,27) = 3}}}. So, the found solution x =15 is correct. </pre> <H3>Exercise 1</H3>Solve yourself an equation {{{root(6,x-3) = 2}}}. <H3>Example 2</H3>Solve an equation {{{sqrt(x-1)}}} + {{{sqrt(2x+6)}}} = {{{6}}}. <B>Solution</B> <pre> Let us isolate the radical expression {{{sqrt(2x+6)}}} on the left side: {{{sqrt(2x+6)}}} = {{{6 - sqrt(x-1)}}}. Now raise both sides to a degree 2 (the index of the radical is 2). You get {{{(sqrt(2x+6))^2}}} = {{{(6 - sqrt(x-1))^2}}}, 2x+6 = {{{36 - 2*6*sqrt(x-1) + x-1}}}, and after simplifications {{{12*sqrt(x-1) = 29-x}}}. Since the equation still contains radical (now on the left side only), raise both sides to a degree 2 one more time: 144*(x-1) = {{{(29-x)^2}}}. Open parentheses, collect all common terms on one side. You get 144x-144 = {{{841 - 58x +x^2}}}, {{{x^2 - 202x + 985}}} = {{{0}}}. The last equation is a quadratic equation. Apply quadratic formula (see the lesson <A HREF = http://www.algebra.com/algebra/homework/quadratic/lessons/Introduction-Into-Quadratics.lesson>Introduction into Quadratic Equations</A>) {{{x[1,2]}}} = {{{(202+-sqrt(202^2-4*985))/2}}} = {{{(202+-sqrt(40804-3940))/2}}} = {{{(202 +-sqrt(36864))/2}}}. Two roots of this equation are {{{x[1] = 5}}} and {{{x[2] = (202+192)/2 = 197}}}. To check it, simply substitute these values into the original equation. For {{{x=5}}} you have {{{sqrt(5-1)+sqrt(2*5+6)}}} = {{{sqrt(4)+sqrt(16)}}} = 2+4 = 6. This means {{{x=5}}} is the solution of the original equation. For {{{x=197}}} you have {{{sqrt(197-1)+sqrt(2*197+6)}}} = {{{sqrt(196)+sqrt(400)}}} = 14+20 = 34. It is not equal to 6, so x=197 is not a solution. This is an <B>extraneous solution</B>. <B>Answer</B>. The solution is x=5. </pre> <H3>Exercise 2</H3>Solve yourself an equation {{{sqrt(2x+3) - sqrt(x+2)}}} = {{{2}}}. <H3>Problem 3</H3>Solve an equation {{{sqrt(24-10x)}}} = {{{3-4x}}}, where 3-4x > 0. <B>Solution</B> <pre> 1. Square both sides to get 24 - 10x = {{{(3-4x)^2}}} 24 - 10x = {{{9 - 24x + 16x^2}}} {{{16x^2 - 14x - 15}}} = 0 (8x + 5)*(2x - 3) = 0 The roots are x = {{{-5/8}}} and x = {{{3/2}}}. With x = {{{3/2}}} the expression (24-10x) under the square root is negative and therefore does not work. The only working solution is x = {{{-5/8}}}. See the plot below. {{{graph( 330, 330, -5.5, 5.5, -5.5, 10.5, sqrt(24-10x), 3-4x )}}} Plot y = {{{sqrt(24-10x)}}} (red) and y = 3-4x (green) </pre> <H3>Problem 4</H3>Solve an equation {{{4x - 11*sqrt(x) + 6}}} = {{{0}}}. <B>Solution</B> <pre> Introduce new variable y = {{{sqrt(x)}}}. Then your equation takes the form 4y^2 - 11y + 6 = 0. Solve this quadratic equation using the quadrtic formula {{{y[1,2]}}} = {{{(11 +- sqrt((-11)^2 - 4*4*6))/(2*4)}}} = {{{(11 +- sqrt(25))/8}}} = {{{(11 +- 5)/8}}}. Case 1. {{{y[1]}}} = {{{(11+5)/8}}} = {{{16/8}}} = 2. Then {{{sqrt(x)}}} = 2; hence, x = 4. Case 2. {{{y[2]}}} = {{{(11-5)/8}}} = {{{6/8}}} = {{{3/4}}}. Then {{{sqrt(x)}}} = {{{3/4}}}; hence, x = {{{9/16}}}. <U>ANSWER</U>. The given equation has two and only two solutions: x= 4 and x= {{{9/16}}}. </pre> <H3>Problem 5</H3>Solve an equation {{{z^2-6z+9}}} = {{{4*sqrt(z^2-6z+6)}}}. <B>Solution</B> <pre> {{{z^2-6z+9}}} = {{{4*sqrt(z^2-6z+6)}}} Rewrite equivalently {{{(z-3)^2}}} = {{{4*sqrt((z-3)^2-3)}}}. Square both sides: {{{(z-3)^4}}} = {{{16*((z-3)^2-3)}}}. Introduce new variable x = {{{(z-3)^2}}}. The last equation takes the form {{{x^2}}} = {{{16(x -3)}}}, or {{{x^2 - 16x + 48}}} = {{{0}}}. Solve by any way (factoring or the quadratic formula). You will get the roots x = 12 and/or x = 4. Thus you have two options: 1. {z-3)^2 = 4 ---> z-3 = 2 or z-3 = -2 ---> z = 5 or z = 1. 2. (z-3)^2 = 12 ---> z-3 = +/-{{{sqrt(12)}}} = +/-{{{2*sqrt(3)}}} ---> z = {{{3+2sqrt(3)}}} and/or z = {{{3-2sqrt(3)}}}. Check that all these roots are the solutions of the original equation. <B>Answer</B>. The solutions are z = 1, z = 5, z = {{{3+2sqrt(3)}}}, z = {{{3-2sqrt(3)}}}. </pre> <TABLE> <TR> <TD> {{{graph( 330, 330, -2.5, 8.5, -2.5, 15.5, x^2-6x+9, 4*sqrt(x^2-6x+6) )}}} Plots y = {{{x^2-6x+9}}} (red), y = {{{4*sqrt(x^2-6x+6)}}} (green) </TD> </TR> </TABLE> <H3>Problem 6</H3>Solve an equation {{{root(2/3,x-2) - root(1/3,x-2) = 2}}}. <B>Solution</B> <pre> Let us apply the method that introduces a new variable. Put {{{y=root(1/3,x-2)}}}. Then {{{y^2=root(2/3,x-2)}}}, and you can rewrite the given equation in the form {{{y^2 - y - 2}}} = {{{0}}}. It is a quadratic equation. It has the roots {{{y=2}}} and {{{y=-1}}}. Now the problem is to solve the equations {{{root(1/3,x-2) = 2}}} and {{{root(1/3,x-2) = -1}}}. Raising both sides of the equation {{{root(1/3,x-2) = 2}}} to degree 3 (the index of the radical is 3) you get {{{x-2=2^3=8}}}. Hence, {{{x=10}}}. The second equation {{{root(1/3,x-2) = -1}}} has no solution in real numbers, because the domain for the fractional power is the set of positive real numbers, but the fractional power of the positive real number is the positive real number again. <B>Answer</B>. {{{x=10}}}. </pre> <H3>Exercise 3</H3>Solve yourself the equation {{{root(2/5,x-5)-2*root(1/5,x-5) = 3}}} introducing new variable. My other lesson on solving equations containing radicals is - <A HREF=https://www.algebra.com/algebra/homework/Radicals/Advanced-problems-on-solving-equations-containing-radicals.lesson>Advanced problems on solving equations containing radicals</A> - <A HREF=https://www.algebra.com/algebra/homework/Radicals/Solving-systems-of-equations-containing-radicals.lesson>Solving systems of equations containing radicals</A> - <A HREF=https://www.algebra.com/algebra/homework/Radicals/OVERVIEW-of-my-lessons-on-solving-equations-containing-radicals.lesson>OVERVIEW of my lessons on solving equations containing radicals</A> in this site. Use this file/link <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-I.
