Solve the equation 2x^2 + 8x - 42 = 0. Part B: In this part, the unit of length is the centimeter. ABC is a triangle such that AB = x, AC = x + 4 and BC = sqrt(58), where x is an i... Solve the equation 2x^2 + 8x - 42 = 0. Part B: In this part, the unit of length is the centimeter. ABC is a triangle such that AB = x, AC = x + 4 and BC = sqrt(58), where x is an integer strictly greater than 1. 1) Can we find a value for x such that triangle ABC is right-angled at C? Justify. 2) Calculate x so that triangle ABC is right-angled at A. (You can use the results of part A). 3) Calculate x so that the perimeter of triangle ABC is less than or equal to 18. Read more

Steps to Solve

1. Solve the Quadratic Equation

   First, we need to solve the quadratic equation (2x^2 + 8x - 42 = 0). We can simplify this by dividing the entire equation by 2:

   $$ x^2 + 4x - 21 = 0 $$

   Next, we apply the quadratic formula:

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

   Here, (a = 1), (b = 4), and (c = -21).

2. Calculate the Discriminant

   We calculate the discriminant (D):

   $$ D = b^2 - 4ac = 4^2 - 4(1)(-21) = 16 + 84 = 100 $$

   Since the discriminant is positive, we proceed to find the roots.

3. Find the Roots

   Substitute back into the quadratic formula:

   $$ x = \frac{-4 \pm \sqrt{100}}{2(1)} $$ $$ x = \frac{-4 \pm 10}{2} $$

   This results in two potential values for (x):

   $$ x = \frac{6}{2} = 3 $$ $$ x = \frac{-14}{2} = -7 $$

   We discard (x = -7) since (x) must be greater than 1. Hence, (x = 3).

4. Part B: Right-Angled Triangle Conditions

   For ABC with (AB = x), (AC = x + 4), and (BC = \sqrt{58}), we now check if a right angle can exist at (C):

   Using the Pythagorean theorem:

   $$ AB^2 + AC^2 = BC^2 $$

   Which translates to:

   $$ x^2 + (x + 4)^2 = 58 $$

5. Expand and Simplify

   Expand ((x + 4)^2):

   $$ x^2 + (x^2 + 8x + 16) = 58 $$ $$ 2x^2 + 8x + 16 - 58 = 0 $$ $$ 2x^2 + 8x - 42 = 0 $$

   We have already solved this in step 1 and found (x = 3).

6. Right Angled at A

   Now check if the triangle is right-angled at (A):

   $$ AC^2 + BC^2 = AB^2 $$

   Thus,

   $$ (x + 4)^2 + 58 = x^2 $$

   Substitute (x = 3):

   $$ (3 + 4)^2 + 58 = 3^2 $$ $$ 49 + 58 \neq 9 $$ (not true)

   So, there is no (x) for which triangle ABC is right-angled at (A).

7. Perimeter Condition

   Now ensure the perimeter is ( \leq 18):

   $$ AB + AC + BC \leq 18 $$ $$ x + (x + 4) + \sqrt{58} \leq 18 $$ $$ 2x + 4 + \sqrt{58} \leq 18 $$

   Calculate:

   $$ 2x \leq 18 - 4 - \sqrt{58} $$ $$ 2x \leq 14 - 7.62 \approx 6.38 $$

   Therefore:

   $$ x \leq 3.19 $$

   The only integer greater than 1 is (x = 3).