# AB = 4x - 15 and BC = 2x + 3. If AC = 48. Find the value of x. (AC = hypotenuse )

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Given the line segments AB, BC and AC such that:

AB = 4x - 15

BC = 2x + 3

AC = 48.

Also given that AC is a hypotenuse.

Then BC and AB are the other sides of the right angle triangle.

Then we know that:

AC^2 = AB^2 + BC^2

48^2 = ( 4x-15)^2 + ( 2x+3)^2

2304 = 16x^2 - 180x + 225 + 4x^2 + 12x + 9

2304 = 20x^2 - 169x + 234

==> 20x^2 - 169x - 2070 = 0

==> x = ( 169 + 440.6) / 40

**==> x1=15.24**

==> x2 ( 169 - 440.6) /40

=-6.79

**==>x2= -6.79**

Since the given sides represent the cathetus and the hypothenuse of a right angle triangle, we'll determine x applying Pythagorean theorem:

hypothenuse^2 = cathetus^2 + cathetus^2

From enunciation, the hypothenuse is AC:

AC^2 = AB^2 + BC^2

48^2 = (4x - 15)^2 + (2x + 3)^2

We'll expand the square form the right side:

2304 = 16x^2 - 120x + 225 + 4x^2 + 12x + 9

We'll combine like terms and we'll use symmetric property:

20x^2 - 108x - 2070 = 0

10x^2 - 54x - 1035 = 0

We'll apply the quadratic formula:

x1 = [54+sqrt(44316 )]/20

x1 = (54+210.51)/20

**x1 = 13.2255 approx.**

**x2 = -7.8255 approx.**

AB =4x-15 and BC = 2x+3. AC = 48. AC is the hypotenuse.

To find the value of x.

From the given details, ABC is a right angled triangle with AC as hypotenuse . Therefore AB and AC form a right angle at B.

So AB^2+BC^2 = AC^2 ....(1) by Pythagoas theorem.

We substitute the given values AB = 4x - 15 and BC = 2x + 3 and AC = 48 in (1) and solve the quadratic equation for x. :

(4x-15)^2 +(2x+3)^2 = 48^2.

16x^2 -120x +225 +4x^2+12x +9 - 48^2 = 0

20x^2-108x - 2070 = 0....(1)

We use the quadratic formula : The solution of ax^2+bx+c= 0 is x1 = {-b +or- sqrt(b^2-4ac)}/2a. Here a= 20, b = -108 and c = 2070.

So x1 = {108 +sqrt(108^2-4*20(-2070)}/2*40 = 13.2257 nearly.

x2 = {108 +sqrt(108^2-4*20(-2070)}/2*40 = -7.8257 nearly.