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Evaluate:
 $ x(2x + 5) = 25 $

Answer
VerifiedVerified
521.4k+ views
Hint: First of all multiply the term with all the terms inside the bracket, move all the terms on one side of the equation, if you move any term from one side to another then the sign of the terms also changes. Then will factorize to find the required value for “x”.

Complete step-by-step answer:
Take the given expression: $ x(2x + 5) = 25 $
Multiply the term outside the bracket with all the terms inside the bracket.
 $ x(2x) + x(5) = 25 $
Simplify the above expression finding the product of the terms.
 $ 2{x^2} + 5x = 25 $
Move the constant term from the right hand side of the equation to the left hand side of the equation. When you move any term from one side to the opposite side then the sign of the term also changes. Positive term changes to negative and vice-versa.
 $ 2{x^2} + 5x - 25 = 0 $
Split the middle term in such a way that the product of the terms is equal to the product of the first and the last term.
For example:
 $
   + 5 = 10 - 5 \\
  2( - 25) = - 50 = (10)( - 5) \;
  $
Now split the middle term and re-write the equation –
 $ 2{x^2} + \underline {10x - 5x} - 25 = 0 $
Make the pair of first term and the last term
 $ \underline {2{x^2} + 10x} - \underline {5x - 25} = 0 $
Take common multiples from both the paired terms –
 $ 2x(x + 5) - 5(x + 5) = 0 $
Now, write common multiple common
 $ (x + 5)(2x - 5) = 0 $
We get two values for “x”
 $
  x + 5 = 0 \\
   \Rightarrow x = ( - 5) \;
  $
And $ (2x - 5) = 0 $
Simplify –
 $ 2x = 5 $
Term multiplicative on one side if moved to the opposite side then it goes in the denominator.
 $ x = \dfrac{5}{2} $
Hence, the required values for “x” are $ x = ( - 1) $ and $ x = \dfrac{5}{2} $
So, the correct answer is “ $ x = ( - 1) $ and $ x = \dfrac{5}{2} $ ”.

Note: Be careful about the sign convention and splitting the middle term. The product of splits should be equal to the product of the first and the last term along with its sign. Remember the product of negative term and the positive term is negative term.
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