One leg of a right triangle is 7 inches shorter than the other leg. When is the hypotenuse at least 17 inches long?(1 point)

Let's denote the length of the longer leg as \( x \) inches. According to the problem, the shorter leg will then be \( x - 7 \) inches.

We can use the Pythagorean theorem, which states that for a right triangle with legs \( a \) and \( b \), and hypotenuse \( c \),

\[ c^2 = a^2 + b^2. \]

In our case, we have:

\[ c = \text{hypotenuse} \quad (c \geq 17 \text{ inches}), \] \[ a = x - 7, \quad b = x. \]

Thus, the equation becomes:

\[ c^2 \geq (x - 7)^2 + x^2. \]

We need to find when \( c \) is at least 17 inches, so we set up the inequality:

\[ 17^2 \leq (x - 7)^2 + x^2. \]

Calculating \( 17^2 = 289 \), we have:

\[ 289 \leq (x - 7)^2 + x^2. \]

Expanding \( (x - 7)^2 \):

\[ (x - 7)^2 = x^2 - 14x + 49. \]

Putting this in the inequality, we have:

\[ 289 \leq (x^2 - 14x + 49) + x^2, \] \[ 289 \leq 2x^2 - 14x + 49. \]

Bringing all terms to one side gives:

\[ 0 \leq 2x^2 - 14x + 49 - 289, \] \[ 0 \leq 2x^2 - 14x - 240. \]

Dividing the whole inequality by 2 for simplicity:

\[ 0 \leq x^2 - 7x - 120. \]

To solve this quadratic inequality, we first find the roots of the equation \( x^2 - 7x - 120 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Here, \( a = 1 \), \( b = -7 \), and \( c = -120 \):

\[ x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-120)}}{2 \cdot 1}, \] \[ x = \frac{7 \pm \sqrt{49 + 480}}{2}, \] \[ x = \frac{7 \pm \sqrt{529}}{2}, \] \[ x = \frac{7 \pm 23}{2}. \]

Calculating the roots:

1. \( x = \frac{30}{2} = 15 \)
2. \( x = \frac{-16}{2} = -8 \)

Thus, the roots \( x = 15 \) and \( x = -8 \) define critical points.

To find the intervals for which \( x^2 - 7x - 120 \geq 0 \), we test intervals around the critical points \( (-\infty, -8) \), \( (-8, 15) \), and \( (15, \infty) \):

• For \( x \) in \( (-\infty, -8) \) and \( (15, \infty) \), the expression is positive.
• For \( x \) in \( (-8, 15) \), the expression is negative.

Thus, the solution to the inequality \( x^2 - 7x - 120 \geq 0 \) is:

\[ x \leq -8 \quad \text{or} \quad x \geq 15. \]

Given the context of the problem where the legs of the triangle must be positive, we disregard the interval \( x \leq -8 \) (as lengths cannot be negative). Consequently, we conclude:

When the longer leg is greater than or equal to 15 inches.