Question

Find the value of x.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The correct option is B.

Step-by-step explanation:

From the given figure it is clear that triangle LMN is a right angled triangle.

According to Pythagoras theorem: In a right angled triangle

$hypotenuse^2=perpendicular^2+base^2$

In triangle LMN,

$LN^2=LM^2+MN^2$

$(\sqrt{20})^2=x^2+(x-2)^2$

$20=x^2+x^2-4x+4$

$20=2x^2-4x+4$

Subtract 20 from both sides.

$0=2x^2-4x+4-20$

$0=2x^2-4x-16$

Taking out common factors.

$0=2(x^2-2x-8)$

Splitting the middle term we get

$0=2(x^2-4x+2x-8)$

$0=2(x(x-4)+2(x-4))$

$0=2(x-4)(x+2)$

Using zero product property we get

$x-4=0\Rightarrow x=4$

$x+2=0\Rightarrow x=-2$

The value of can not be negative because side of a triangle is always positive. So, the value of x is 4.

Therefore the correct option is B.