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Section 3.1 Extraction of Roots

Subsection 1. Evaluate quadratic expressions

When squaring a negative number, don't forget to enclose it in parentheses. For example, if \(x=-4\text{,}\) then

\begin{equation*} x^2=(-4)^2 = (-4)(-4) = 16 \end{equation*}

If we write \(-4^2\text{,}\) then only the 4 is squared, so we have

\begin{equation*} -4^2 = - (4)(4) = -16 \end{equation*}

Subsubsection Examples

Example 3.1.

Evaluate for \(x=-6\text{.}\)

  1. \(\displaystyle 2x^2\)
  2. \(\displaystyle 2-x^2\)
  3. \(\displaystyle (2x)^2\)
  4. \(\displaystyle (2-x)^2\)
Solution

Enclose \(-6\) in parentheses, and follow the order of operations.

  1. Square first, then multiply by 2: \(~~2x^2=2(-6)^2=2(36)=72\)
  2. Square first, then subtract from 2: \(~~2-x^2=2-(-6)^2=2-36=-34\)
  3. Multiply by 2 first, then square: \(~~(2x)^2=[2(-6)]^2=[-12]^2=144\)
  4. Subtract from 2 first, then square: \(~~(2-x)^2=[2-(-6)]^2=[8]^2=64\)
Example 3.2.

Make a table of values for \(y=x^2+2x-8\text{,}\) and graph the equation.

Solution

We plot the points from the table and connect them with a smooth curve.

\(x\) \(y\) \(\hphantom{0000}\)
\(-5\) \(7\) \((-5)^2+2(-5)-8=25-10-8\)
\(-4\) \(0\) \((-4)^2+2(-4)-8=16-8-8\)
\(-3\) \(-5\) \((-3)^2+2(-3)-8=9-6-8\)
\(-2\) \(-8\) \((-2)^2+2(-2)-8=4-4-8\)
\(-1\) \(-9\) \((-1)^2+2(-1)-8=1-2-8\)
\(0\) \(-8\) \((0)^2+2(0)-8=0+0-8\)
\(1\) \(-5\) \((1)^2+2(1)-8=1+2-8\)
\(2\) \(0\) \((2)^2+2(2)-8=4+4-8\)
\(3\) \(7\) \((3)^2+2(3)-8=9+6-8\)

parabola y=xsqd+2x-8

Subsubsection Exercises

Evaluate for \(w=-9\)

  1. \(\displaystyle (2w)^2\)
  2. \(\displaystyle 36-(2w)^2\)
  3. \(\displaystyle -2(4-w)^2\)
  4. \(\displaystyle 2-w^2\)
Answer
  1. \(\displaystyle 324\)
  2. \(\displaystyle -288\)
  3. \(\displaystyle -338\)
  4. \(\displaystyle -79\)

Evaluate for \(a=-3,~b=-4\)

  1. \(\displaystyle ab^2\)
  2. \(\displaystyle a-b^2\)
  3. \(\displaystyle (a-b^2)^2\)
  4. \(\displaystyle ab(a^2-b^2)\)
Answer
  1. \(\displaystyle -48\)
  2. \(\displaystyle -19\)
  3. \(\displaystyle 361\)
  4. \(\displaystyle -84\)

Evaluate for \(h=-2,~g=-5\)

  1. \(\displaystyle h^2-2hg+g^2\)
  2. \(\displaystyle (h-g)^2\)
  3. \(\displaystyle h^2-g^2\)
  4. \(\displaystyle (h-g)(h+g)\)
Answer
  1. \(\displaystyle 9\)
  2. \(\displaystyle 9\)
  3. \(\displaystyle -21\)
  4. \(\displaystyle -21\)

Make a table of values for \(y=x^2-x-6\text{,}\) and graph the equation.

\(x\) \(y\)
\(-3\) \(\hphantom{0000}\)
\(-2\) \(\hphantom{0000}\)
\(-1\) \(\hphantom{0000}\)
\(0\) \(\hphantom{0000}\)
\(1\) \(\hphantom{0000}\)
\(2\) \(\hphantom{0000}\)
\(3\) \(\hphantom{0000}\)
\(4\) \(\hphantom{0000}\)
\(5\) \(\hphantom{0000}\)

grid
Answer

\(x\) \(y\)
\(-3\) \(6\)
\(-2\) \(0\)
\(-1\) \(-4\)
\(0\) \(-6\)
\(1\) \(-6\)
\(2\) \(-4\)
\(3\) \(0\)
\(4\) \(6\)
\(5\) \(14\)

parabola

Subsection 2. Use square roots

The simplest quadratic equations have the form

\begin{equation*} ax^2+c=0 \end{equation*}

They are missing a linear term, \(bx\text{,}\) and are not difficult to solve.

Subsubsection Examples

Example 3.7.

Solve \(~3y^2-40=35~\) by extraction of roots.

Solution

We isolate the quadratic term, \(y^2\text{,}\) and then extract roots.

\begin{align*} 3y^2-40 \amp = 35 \amp \amp \blert{\text{Add 40 to both sides.}}\\ 3y^2 \amp = 75 \amp \amp \blert{\text{Divide both sides by 3.}}\\ y^2 \amp = 25 \amp \amp \blert{\text{Take the square root of both sides.}}\\ y \amp = \pm \sqrt{25} \amp \amp \blert{\text{Simplify.}}\\ y \amp = \pm 5 \amp \amp \blert{\text{Remeber that a positive number has two square roots.}} \end{align*}

The solutions are \(5\) and \(-5\text{.}\)

Example 3.8.

A cone is 16 cm tall and its volume is 500 cm\(^3\text{.}\) What is the radius of the base of the cone?

Solution

We substitute \(h=16\) and \(V=500\) into the formula for the volume of a cone, and solve for \(r\text{.}\)

\begin{align*} \dfrac{1}{3} \pi r^2(\alert{16}) \amp = \alert{500} \amp \amp \blert{\text{Multiply both sides by 3.}}\\ 16 \pi r^2 \amp = 1500 \amp \amp \blert{\text{Divide both sides by}~16 \pi.}\\ r^2 \amp = 29.842 \amp \amp \blert{\text{Take the square root of both sides.}}\\ r \amp = 5.46 \end{align*}

The radius of the cone is 5.46 cm.

Subsubsection Exercises

Solve by extraction of roots.

  1. \(\displaystyle 98=2a^2\)
  2. \(\displaystyle 0=3n^2-15\)
  3. \(\displaystyle 144 + \dfrac{h^2}{9} = 169\)
  4. \(\displaystyle 12-5v^2=2\)
Solution
  1. \(\displaystyle \pm 7\)
  2. \(\displaystyle \pm \sqrt{5}\)
  3. \(\displaystyle \pm \sqrt{15}\)
  4. \(\displaystyle \pm \sqrt{2}\)

A cylindrical syringe holds 100 cc (cubic centimeters) of fluid. If the syringe is 10 centimeters long, what is its radius?

Solution

1.78 cm

Subsection 3. Use the Pythagorean theorem

If \(a\) and \(b\) are the lengths of the legs of a right triangle and \(c\) is the length of the hypotenuse, then

\begin{equation*} a^2 + b^2 = c^2 \end{equation*}

Note that the theorem is true only for right triangles -- ones that have a 90\(\degree\) angle.

Subsubsection Examples

Example 3.11.

Find the unknown side in the right triangle.

right triangle
Solution

The unknown side is the hypotenuse, so we apply the Pythagorean theorem with \(c=z,~a=12\text{,}\) and \(b=35\text{.}\)

\begin{align*} a^2 + b^2 \amp = c^2\\ 12^2+35^2 \amp =z^2 \amp \amp \blert{\text{Simplify the left side.}}\\ 1369 \amp = z^2 \amp \amp \blert{\text{Take the square root of both sides.}}\\ \pm 37 \amp = z \end{align*}

The length of the hypotenuse is a positive number, so \(z=37\text{.}\)

Example 3.12.

A 25-foot ladder is placed against a wall so that its foot is 7 feet from the base of the wall. How far up the wall does the ladder reach?

Solution

We make a sketch and label the known dimensions, calling the unknown height \(h\text{.}\) The ladder forms the hypotenuse of a right triangle, so we apply the Pythagorean theorem, substituting 25 for \(c\text{,}\) 7 for \(b\text{,}\) and \(h\) for \(a\text{.}\)

\begin{align*} a^2 + b^2 \amp = c^2\\ h^2 + 7^2 \amp = 25^2 \end{align*}
ladder

We solve the equation by extraction of roots:

\begin{align*} h^2 + 49 \amp = 625 \amp \amp \blert{\text{Subtract 49 from both sides.}}\\ h^2 \amp =576 \amp \amp \blert{\text{Extract roots.}}\\ h \amp =\pm \sqrt{576} \amp \amp \blert{\text{Simplify the radical.}}\\ h = \amp \pm 24 \end{align*}

The height must be a positive number, so the ladder reaches 24 feet up the wall.

Subsubsection Exercises

A baseball diamond is a square whose sides are 90 feet long. Find the straight-line distance from home plate to second base.

  1. Make a sketch of the situation and label a right triangle.
  2. Write an equation and solve.
Answer
  1. baseball diamond
  2. 127.28 ft

How long a wire is needed to stretch from the top of a 40-foot telephone pole to a point on the ground 20 feet from the base of the pole?

  1. Make a sketch of the situation and label a right triangle.
  2. Write an equation and solve.
Answer
  1. telephone pole and wire
  2. 44.72 ft