Question

How do you complete the square to solve $ - {x^2} - 2x + 3 = 0$?

Answer 1

Hint: First multiply the terms by $ - 1$. After that, we will be applying the completing square method to find $x$ . We complete the square by adding the same numbers to both sides of the quadratic equation and then try to get it in a square form.

Complete step-by-step answer:
Before proceeding with the question, we should know about completing the square method.
To solve the quadratic equation of form $a{x^2} + bx + c = 0$ by completing the square:
(i) Transform the quadratic equation so that the constant term $c$ is alone on the right side.
(ii) If $a$ , the leading coefficient is not equal to 1, divide both sides by $a$.
(iii) Add the square of half the coefficient of $x$ term, ${\left( {\dfrac{b}{{2a}}} \right)^2}$ to both sides of the equation.
(iv) Factor the left side as the square of a binomial.
(v) And now solve for $x$.
So now following the above-mentioned method we get,
$ \Rightarrow - {x^2} - 2x + 3 = 0$
Multiply both sides by $ - 1$,
$ \Rightarrow {x^2} + 2x - 3 = 0$
Adding 3 on both sides of the equation we get,
$ \Rightarrow {x^2} + 2x = 3$
Here coefficient: $a = 1,b = 2,c = - 3$. Now adding ${\left( 1 \right)^2}$ to both sides of the equation we get,
$ \Rightarrow {x^2} + 2x + {\left( 1 \right)^2} = 3 + {\left( 1 \right)^2}$
Solving both sides of the equation, we get
$ \Rightarrow {x^2} + 2x + 1 = 3 + 1$
Using the identity, ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$, the above equation will be,
$ \Rightarrow {\left( {x + 1} \right)^2} = 4$
Taking the square root of both the sides of the equation we get,
$ \Rightarrow x + 1 = \pm 2$
Move 1 to the right side,
$ \Rightarrow x = - 1 - 2, - 1 + 2$
Simplify the terms,
$ \Rightarrow x = - 3,1$

Hence, the value of $x$ is -3 and 1.

Note:
 In this type of question where it is mentioned that we use the square method to solve, we have to memorize the steps involved in the method. We can make mistakes if we do not know that the square root will give two values one positive and one negative.