Question

1. a ladder is resting against a wall. the top of the ladder touches the wall at a height of 15 feet. find the distance from the wall to the bottom of the ladder if the length of the ladder is one foot more than twice its distance from the wall.

Explanation:

Step1: Set up variables

Let the distance from the wall to the bottom of the ladder be \(x\) feet. Then the length of the ladder is \((2x + 1)\) feet. Using the Pythagorean theorem (since the ladder, wall, and ground form a right - triangle with the ladder as the hypotenuse and the height on the wall \(h = 15\) feet), we have \(x^{2}+15^{2}=(2x + 1)^{2}\).

Step2: Expand the equation

Expand \((2x + 1)^{2}\) using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), where \(a = 2x\) and \(b = 1\). So \(x^{2}+225=4x^{2}+4x + 1\).

Step3: Rearrange the equation

Move all terms to one side: \(4x^{2}+4x + 1-x^{2}-225 = 0\). Combine like terms to get \(3x^{2}+4x-224 = 0\).

Step4: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 3\), \(b = 4\), \(c=-224\)), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). First, calculate the discriminant \(\Delta=b^{2}-4ac=(4)^{2}-4\times3\times(-224)=16 + 2688 = 2704\). Then \(\sqrt{\Delta}=\sqrt{2704}=52\). So \(x=\frac{-4\pm52}{6}\). We have two solutions for \(x\): \(x_{1}=\frac{-4 + 52}{6}=\frac{48}{6}=8\) and \(x_{2}=\frac{-4-52}{6}=\frac{-56}{6}\) (reject the negative solution since distance cannot be negative).

Answer:

The distance from the wall to the bottom of the ladder is \(8\) feet.