Trigonometry: How do I solve for X in the following right triangle?

X^2+(X+3)^2=(sqrt17)^2

X^2+X^2+6X+9=17

2X^2+6X-8=0

x^2+3X-4=0

(x+4)(x-1)=0

x=-4............x=1..............cant have a negative side of a triangle

X=1

It's a right triangle so you get to use the Pythagorean Theorem for this one

The sum of the squared sides equals the hypotenuse squared.

x^2 + (x+3)^2 = 17

multiply the second term out using FOIL.

Add like terms.

Subtract 17 from both sides to set your equation equal to zero.

Solve for x just like you would with any 2nd degree polynomial. Disregard any negative solutions.

That is AC's length. Add 3 for CB.

let's compute using the pythagorean theorem which states that

....."the square of the hypotenuse is equal to the sum of the squares of its sides."

AB is the hypotenuse because it is opposite the right angle.

( √ 17 ) ² = x ² + ( x + 3 ) ²

17 = x ² + x ² + 6x + 9

.... = 2x ² + 6x + 9

2x ² + 6x - 8 = 0

( 2x + 8 ) ( x - 1 ) = 0

2x + 8 = 0 ................ & ................ x - 1 = 0

2x = – 8 ................ & ................ x = 1

x = – 4 ................ & ................ x = 1

............................ x cannot be negative, so;

x = 1

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