How do you solve the equation by completing the square x2-14x+33=0? - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

Let's solve this question by the complete square method.
⇒

x^{2} - 14 x + 33 = 0

⇒

x^{2} - 14 x = 33

………………1
⇒ Third term =

{(\frac{1}{2} \times 12)}^{2}

= 49
Adding the term on both sides

\Rightarrow x^{2} - 14 x + 49 = - 33 + 49

\Rightarrow x^{2} - 14 x + 49 = 16

\Rightarrow

(x - 7)^{2} = 16

Taking square roots on both sides

\Rightarrow \sqrt{(x - 7)^{2}} = \sqrt{16}

\Rightarrow

(x - 7)^{} = \pm 4

………………… Let's, find the values of

x

by equating the expression obtained in the solution to the +4 and −4.

\Rightarrow (x - 7) = + 4

\Rightarrow x = 4 + 7

\Rightarrow x = 11

Also,

\Rightarrow (x - 7) = - 4

\Rightarrow x = - 4 + 7

\Rightarrow x = 3

How do you solve the equation by completing the square $x^{2} - 14 x + 33 = 0$ ?