Linear equation with radical expression

Linear Equation with radical expression are explained with the help of this question.

Q# 1: Solve $\sqrt{2x+11}=\sqrt{3x+7}$

Solution:

• First of all just write the equation,

  $\sqrt{2x+11}=\sqrt{3x+7}$

• Then remove the roots from both side by taking square on both side.

  Squaring on both side:

  $(\sqrt{2x+11})^{2}=(\sqrt{3x+7})^{2}$

  Note: Only take square when both side are simplified and their is nothing else to do then taking square.

• The whole square will cancel the roots

  ${2x+11}={3x+7}$

• Now it is a simple linear equation and solve it the same way.

  $11-7=3x-2x$

  $4=x$

  $x=4$

• Now, when solving radical equation the root is not always the solution set so before making a solution set, First verify all the solution

  Verification:

  Put x=4 in the equation,

  $\sqrt{2(4)+11}=\sqrt{3(4)+7}$

  $\sqrt{8+11}=\sqrt{12+7}$

  $\sqrt{19}=\sqrt{19}$

  Thus the solution set is {4}.

Q # 2: Solve $10\sqrt{x+20}=100$

Solution:

• First of all just write the equation,

  $10\sqrt{x+20}=100$

• Then before removing the roots from both side just solve the equation first by dividing both side by 10

  $\frac{10\sqrt{x+20}}{10}=\frac{100}{10}$

  $\sqrt{x+20}=10$

• Now solve it as same as other.

  $(\sqrt{x+20})^{2}=(10)^{2}$

  $x+20=100$

  $x=100-20$

  $x=80$

  Verification:

  Put 80 in equation,

  $10\sqrt{80+20}=100$

  $10\sqrt{100}=100$

  $10(10)=100$

  $100=100$

  Thus the solution set is $\{80\}$