Question

How do you solve $ \sqrt {x + 5} = 3x - 7 $ and find any extraneous equation ?

Answer 1

Hint: In order to solve and write the expression into the simplest form . The square root is related to figuring out what should be the number which when multiplied by itself is equal to the number under the square root symbol $ \sqrt {} $ . This symbol is known as radical . Since in our case we have given the question in which we have to solve and find the value of x , we will first get rid of the radical and remove the square root as we want the original value of x . For this we will do squaring and simplify the equation . Furthermore , by using the quadratic formula we will find the required solution .

Complete step-by-step answer:
If we see the question , we need to solve the given expression under square root which is $ \sqrt {x + 5} = 3x - 7 $ . -----equation 1
By applying the concept of equivalent equation , we will first do squaring both the sides on $ \sqrt {x + 5} = 3x - 7 $ both the L . H . S . and the R . H . S . as follows –
Also , Remember that the solution should remain within the set of numbers so that it there shouldn’t be a negative value under square root .
  $ \sqrt {x + 5} = 3x - 7 $
 $
  {\left( {\sqrt {x + 5} } \right)^2} = {\left( {3x - 7} \right)^2} \\
  x + 5 = {\left( {3x - 7} \right)^2} \;
  $
Furthermore , we need to simplify and expand the terms so that it becomes the quadratic equation .
 $ x + 5 = 9{x^2} - 42x + 49 $
Now we need to find the zeros so we will rearrange and write it sophistically ,
 $ 0 = 9{x^2} - 43x + 44 $
As the quadratic equation cannot be factorized , we will use the quadratic formula ,
a becomes 9
b becomes -43
And c becomes 44
 $ x= \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Since, $ {b^2} - 4ac = D $
 $ x = \dfrac{{ - b \pm \sqrt D }}{{2a}} $
 $
  D = \sqrt {{{\left( {43} \right)}^2} - 4(9)(44)} \\
  D = \sqrt {265} \;
  $
 $
   \Rightarrow {x_1} = \dfrac{{ - ( - 43) + \sqrt {265} }}{{2(9)}},{x_2} = \dfrac{{ - ( - 43) - \sqrt {265} }}{{2(9)}} \\
    \\
   \Rightarrow {x_1} = \dfrac{{(43) + \sqrt {265} }}{{18}},{x_2} = \dfrac{{(43) - \sqrt {265} }}{{18}} \;
  $
 $ x \approx + 3.293 and 1.485 $ To 3 decimal places .
All values of $ x < - 5 $ are extraneous .
 $ {x_1} = \dfrac{{(43) + \sqrt {265} }}{{18}},{x_2} = \dfrac{{(43) - \sqrt {265} }}{{18}} $ are as exact values .
So, the correct answer is “ $ {x_1} = \dfrac{{(43) + \sqrt {265} }}{{18}},{x_2} = \dfrac{{(43) - \sqrt {265} }}{{18}} $ ”.

Note: Remember for the solution should remain within the set of numbers so that it there shouldn’t be a negative value under square root .
In an equivalent equation which has an identical solution we can raise the same odd power to both L.H.S. and R.H.S. of an equation .
Every Quadratic equation has 2 roots.
Discriminant: $ D = {b^2} - 4ac $
Using Discriminant, we can find out the nature of the roots
If D is equal to zero, then both of the roots will be the same and real.
If D is a positive number then, both of the roots are real solutions.
If D is a negative number, then the root are the pair of complex solutions
Cross check the answer and always keep the final answer simplified .