Question

Solve for x............................

Answer 1

Answer:

$x= 2+\sqrt{30}$

Step-by-step explanation:

Given triangle is right angle triangle

Hypotenuse is x+6

side a= x+3

side b= x+1

apply Pythagorean theorem

$c^2= a^2 + b^2$

$(x+6)^2=(x+3)^2+(x+1)^2$

(x+6)^2 =(x+6)(x+6)= x^2+12x+36

(x+1)^2 = x^2 + 2x+1

(x+3)^2=x^2+6x+9

$x^2+12x+36=x^2 + 2x+1+x^2+6x+9$

$x^2+12x+36=2x^2+8x+10$

Subtract 2x^2 , 8x  and 10 on both sides

$-x^2+4x+26=0$

now apply quadratic formula

a=-1, b= 4  and c=26

$x=\frac{-b+-\sqrt{b^2-4ac}}{2a}$

$x=\frac{-4+-\sqrt{(4)^2-4(-1)(26)}}{2(-1)}$

$x=\frac{-4+-\sqrt{120}}{-2}$

sqrt(120)= 2sqrt(30)

$x=\frac{-4+-2\sqrt{30}}{-2}$

make two equations and solve for x

$x=\frac{-4+2\sqrt{30}}{-2}$  

$x= 2-\sqrt{30}$

 $x=\frac{-4-2\sqrt{30}}{-2}$

$x= 2+\sqrt{30}$