Question

Solve by completing the squares x²-2root5x+1=0

Answer 1

Answer:

x = (√5+2) and (√5-2)

Step-by-step explanation:

Given a quadratic equation such that,

${x}^{2} - 2 \sqrt{5} x + 1 = 0$

To solve this by completing the square method.

$= > {(x)}^{2} - 2(x)( \sqrt{5} ) + {( \sqrt{5} )}^{2} + 1 - {( \sqrt{5} )}^{2} = 0 \\ \\ = > {(x - \sqrt{5} )}^{2} + 1 - 5 = 0 \\ \\ = > {(x - \sqrt{5} )}^{2} - 4 = 0 \\ \\ = > {(x - \sqrt{5} )}^{2} - {(2)}^{2} = 0 \\ \\ = > (x - \sqrt{5} + 2)(x - \sqrt{5} - 2) = 0$

Now, we have two cases.

Case 1.

When (x -√5 + 2) = 0

=> x = (√5 - 2)

Case 2.

When (x- √5 - 2) = 0

=> x = (√5 + 2)

Hence, the solutions are (√5+2) and (√5-2).

Concept Map :-

• ${x}^{2} - 2xy + {y}^{2} = {(x - y)}^{2}$
• ${x}^{2} - {y}^{2} = (x + y)(x - y)$