Question

How do you solve $ \sqrt {10x + 6} - \sqrt {x + 1} = \sqrt {8x - 8} $ ?

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 , by somewhere using equivalent equations . Equivalent equations are said to be algebraic equations that may have the same solutions if we add or subtract the same number to both sides of an equation - Left hand side or Right hand side of the equal to sign . Or we can multiply or divide the same number to both sides of an equation - Left hand side or Right hand side of the equal sign . . After eliminating the radical , we will then solve the equation by applying the identity and find the value of the x .

Complete step-by-step answer:
If we see the question , we need to solve the given expression under square root which is $ \sqrt {10x + 6} - \sqrt {x + 1} = \sqrt {8x - 8} $ . -----equation 1
By applying the concept of equivalent equation , we will first do squaring both sides on $ \sqrt {10x + 6} - \sqrt {x + 1} = \sqrt {8x - 8} $ both the L . H . S . and the R . H . S . as follows –
We are going to isolate a square root on the L . H . S . So that we can apply the identity as stated below –
We can apply the formula or we can say identity $ {(a - b)^2} = {a^2} + {b^2} - 2ab $
 $ \sqrt {10x + 6} - \sqrt {x + 1} = \sqrt {8x - 8} $
 $ {\left( {\sqrt {10x + 6} - \sqrt {x + 1} } \right)^2} = {\left( {\sqrt {8x - 8} } \right)^2} $
 $
  10x + 6 + x + 1 - 2\sqrt {(10x + 6)(x + 1)} = 8x - 8 \\
  11x + 7 - 2\sqrt {(10x + 6)(x + 1)} = 8x - 8 \\
  3x + 15 = 2\sqrt {(10x + 6)(x + 1)} \;
  $
Now , we will do squaring on both the sides L . H . S . and the R . H . S . , we get –
 $
  {(3x + 15)^2} = 4((10x + 6)(x + 1)) \\
  9{x^2} + 90x + 225 = 4(10{x^2} + 16x + 6) \\
  9{x^2} + 90x + 225 = 40{x^2} + 64x + 24 \\
  31{x^2} - 26x - 201 = 0 \;
  $
Here , after expanding and simplifying we got the simplified quadratic equation , we will now solve further using quadratic formula , we get –
a becomes 31
b becomes -26
And c becomes -201
Let’s now find the determinant value of this quadratic equation, as by looking at the value of determinant we can find out the nature of both the roots .
 $ D = {b^2} - 4ac $
Now using Quadratic formula to find out the actual value of the roots
x= $ \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
= $ \dfrac{{26 \pm 160}}{{62}} $
$ {x_1} = 3,{x_2} = - 2.16 $
Therefore , the final and required answer is x=3 .
So, the correct answer is “ x=3 ”.

Note: Always remember the algebraic identities .
Always try to get rid of the square root .
We can use prime factorisation for the number inside the radical and pull out non- radical terms or perfect squares from the inside of the square root to make the solution easier .
In equivalent equations which have identical solutions we can perform multiplication or division by the same non-zero number both L.H.S. and R.H.S. of an equation .
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 .
Cross check the answer and always keep the final answer simplified .