How do you solve $ {x^2} - 11x + 19 = - 5 $ ?

Hint: Solving an equation means to calculate its roots. So, first we have to try to bring the given equation in the standard form of quadratic equations i.e., $ a{x^2} + bx + c = 0 $ . As it is a quadratic equation, we will obtain two roots i.e., two values of $ x $ for which this equation equates to zero. So, once we get the equation in the standard form, we can apply the direct formula to obtain the roots of this equation i.e., $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ . Applying this formula, we will get two values of $ x $ which are called the two roots of the equation.

Complete step-by-step answer:
(i)
We are given:
  $ {x^2} - 11x + 19 = - 5 $
To bring it in the form of $ a{x^2} + bx + c = 0 $ , we will shift $ - 5 $ to the LHS. As we shift $ - 5 $ to LHS, it will become $ + 5 $ and we will get:
  $ {x^2} - 11x + 19 + 5 = 0 $
On further simplifying, we will get:
  $ {x^2} - 11x + 24 = 0 $
(ii)
Now, as we have our equation in the form of $ a{x^2} + bx + c = 0 $ , we can directly apply the formula for getting the roots of the above quadratic equation. The formula we will be using is:
  $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Since, our equation is:
  $ {x^2} - 11x + 24 = 0 $
As we compare the above equation with $ a{x^2} + bx + c = 0 $ , we can say that:
  $
  a = 1 \\
  b = - 11 \\
  c = 24 \;
   $
Now, since we know the value of $ a $ , $ b $ and $ c $ , we will put them in the formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ to obtain the roots:
  $ x = \dfrac{{ - ( - 11) \pm \sqrt {{{( - 11)}^2} - 4(1)(24)} }}{{2(1)}} $
On simplifying it further, we will get:
  $
  x = \dfrac{{11 \pm \sqrt {121 - 96} }}{2} \\
  x = \dfrac{{11 \pm \sqrt {25} }}{2} \;
   $
As we know that $ {5^2} = 25 $ i.e., $ \sqrt {25} = 5 $ , using this in the equation, we will get:
  $ x = \dfrac{{11 \pm 5}}{2} $
(iii)
As we know that the symbol of $ \pm $ indicates that there are two solutions of quadratic equations, one with $ + $ sign, and the other with $ - $ sign. Therefore, we will separate $ x = \dfrac{{11 \pm 5}}{2} $ into two different equations.
Therefore, our first equation will be:
  $ x = \dfrac{{11 + 5}}{2} $
Solving it further, we will get:
  $ x = \dfrac{{16}}{2} $
Therefore, we will get:
  $ x = 8 $
(iv)
Solving the second equation:
  $ x = \dfrac{{11 - 5}}{2} $
Solving it further,
  $ x = \dfrac{6}{2} $
Therefore,
  $ x = 3 $
Since, we have got two values of $ x $ , i.e., $ x = 8 $ and $ x = 3 $ , we can say that the solution of $ {x^2} - 11x + 19 = - 5 $ is $ x = 8 $ and $ x = 3 $ .
So, the correct answer is “ $ x = 8 $ and $ x = 3 $ ”.

Note: If someone forgets the formula of calculating the roots of a quadratic equation, this question can also be solved by splitting the middle term method. In this method, our aim is to write the quadratic equation as a product of two-line quadratic polynomials. This is done by splitting the middle term i.e., $ bx $ from $ a{x^2} + bx + c = 0 $ in two terms such that their sum is $ bx $ and their product is $ ac{x^2} $ . As we split the middle term, the expression becomes easier to factorize and then we can separately calculate the solutions of both the factors which themselves are linear polynomials.