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How do you complete the square to solve $x^{2} - 4 x - 5 = 0$ ?

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An Intiative by Sri Chaitanya

(2, 1)

(5, - 1)

(- 1, 0)

(4, 2)

answer is B.

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Detailed Solution

\Rightarrow (x^{2} - 4 x) - 5 = 0

Taking equation in the form

a x^{2} + bx + c = 0

Where,

a = 1

and

b = - 4

\Rightarrow \frac{- 4}{2} = - 2

(divide

' b'

2

)

\Rightarrow {(- 2)}^{2} = 4

(squared)

\Rightarrow (x^{2} - 4 x + 4) - 5 = 0

(inside the parenthesis)

\Rightarrow (x^{2} - 4 x + 4) - 5 - 4 = 0

(as we added

4

so we will subtract the

4

also)

\Rightarrow (x^{2} - 4 x + 4) - 9 = 0

Here,

a = x

and

b = - 2

{(a + b)}^{2}

\Rightarrow {(x - 2)}^{2} - 9 = 0

\Rightarrow {(x - 2)}^{2} = 9

(Adding

9

on both sides)

\Rightarrow (x - 2) = \sqrt{9}

\Rightarrow x - 2 = \pm 3

\Rightarrow x - 2 = 3

and

x - 2 = - 3

\Rightarrow x = 3 + 2

and

x = - 3 + 2

\Rightarrow x = 5

and

x = - 1

∴ x = 5

and

x = - 1

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