The problem is, triangle ABC is an isosceles triangle where <B is it’s vertex angle. Given AB= x^2+2, BC=15, And AC=x. Find the value(s) Of x and determine each side length(s) Of the triangle.

Not sure if I’ve done it right but I created a picture and set x^2+2x=15 then factored to get x=3, -5. I plugged the 3 back into x for x^2+2x to get AB=15, BC=15, And AC =3. Is this correct?

3 answers

Looks right to me.

Thanks

To solve the problem, you need to use the properties of an isosceles triangle. In an isosceles triangle, two sides have equal lengths. In this case, you are given that AB and BC have different lengths, so it cannot be an isosceles triangle.

Let's review the problem and solve it step by step:

Given:
AB = x^2 + 2
BC = 15
AC = x

To determine the value(s) of x and the side lengths, we need to use the properties of the triangle. Since it is not an isosceles triangle, we cannot assume that AB and BC are equal in length.

First, let's set up an equation using the given information. According to the Triangle Inequality Theorem, the sum of any two sides of a triangle must be greater than the third side. So we have:

AB + BC > AC (1)
AC + BC > AB (2)
AB + AC > BC (3)

Substituting the given values:

(x^2 + 2) + 15 > x (1)
x + 15 > (x^2 + 2) (2)
(x^2 + 2) + x > 15 (3)

Now, let's solve the inequalities one by one:

From equation (2):
x + 15 > x^2 + 2

Rearranging the terms:
x^2 - x + 13 < 0

Since it is a quadratic equation, we can solve it using the quadratic formula:
x = (-(-1) ± √((-1)^2 - 4(1)(13))) / (2(1))
x = (1 ± √(-51)) / 2

Since the discriminant (√(-51)) is negative, there are no real solutions for x. Therefore, the equation does not hold for any real value of x.

Hence, the triangle cannot be formed with the given side lengths (AB = x^2 + 2, BC = 15, AC = x).

In conclusion, your initial assumption that it is an isosceles triangle was incorrect. There are no real values of x that satisfy the given values for the sides.