Question

Solve \[8x - \left( {3x + 5} \right) = 6x - 19\] using the quadratic formula.

Answer 1

Hint: Here in order to solve the above given question using quadratic formula we have to find the values of $ a,\;b,\;c $ corresponding to the given question. Then by substituting the values in the above equation we can find the values for $ x $ and thereby solve it.

Quadratic Formula: $ a{x^2} + bx + c = 0 $ Here $ a,\;b,\;c $ are numerical coefficients.
So to solve $ x $ we have: $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $

Complete step-by-step solution:
Given
\[8x - \left( {3x + 5} \right) = 6x - 19\;..............................\left( i \right)\]
Now we need to compare (i) to the general formula and find the values of unknowns. Then we have to use the equation to find $ x $ by substituting all the values needed in it and by that way we can solve the equation\[8x - \left( {3x + 5} \right) = 6x - 19\;\].
So on comparing (i) to the general formula $ a{x^2} + bx + c = 0 $ , we see that the given \[8x - \left( {3x + 5} \right) = 6x - 19\;\]is not in the general form $ a{x^2} + bx + c = 0 $ .
Therefore it cannot be solved using the quadratic formula.
Now solving \[8x - \left( {3x + 5} \right) = 6x - 19\;\]without using the quadratic formula.
Now in order to solve the equation (i) we need to solve for $ x $ .
Such that we have to manipulate the given equation in terms of only $ x $ , which can be achieved by performing different arithmetic operations on both LHS and RHS equally.
So to isolate the $ x $ terms from equation (i) first let’s add $ 5 $ to both LHS and RHS, since adding $ 5 $ to the LHS will isolate the $ x $ terms by canceling the term $ + 5 $ .
Adding $ 5 $ from both LHS and RHS of equation (i), we get:
\[
  8x - 3x - 5 = 6x - 19\; \\
  8x - 3x - 5 + 5 = 6x - 19\; + 5 \\
  8x - 3x = 6x - 14......................\left( {ii} \right)\; \\
 \]
On simplifying (ii) we can write:
\[
  8x - 3x = 6x - 14 \\
  5x = 6x - 14 \\
  5x - 6x = - 14 \\
   - x = - 14 \\
  x = 14.........................\left( {iii} \right) \\
 \]
Therefore on solving \[8x - \left( {3x + 5} \right) = 6x - 19\;\]without using the quadratic formula we get $ x = 14 $ .


Note: Quadratic formula is mainly used in conditions where grouping method cannot be used or when the polynomial cannot be reduced into some general identity. While solving questions with quadratic formula one should always check whether the given expression of the form $ a{x^2} + bx + c = 0 $ or not. If it is not in the form $ a{x^2} + bx + c = 0 $ then the quadratic formula cannot be used to solve it.