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Second-Degree Equations and Inequalities Solved Problems for Intermediate Algebra

By — McGraw-Hill Professional
Updated on Aug 12, 2011

Review the following concepts if needed:

6.1 Solve the following quadratic equations by the factor method. Check your answers.

  1. x2 + x – 12 = 0
  2. Factor the quadratic expression and apply the zero factor property: x2 + x – 12 = (x + 4) (x – 3). Now solve the resulting linear equations.

    The solution set is {–4,3}.

  3. t2 + 15t + 100 = 0
  4. Follow the procedure used in part (a). t2 + 15t– 100 = (t + 20) (t – 5).

    The solution set is {–20,5 }.

  5. w2 – 2w + 1 = 0
  6. The equation has only one solution, since the expression has only one factor. The solution set is {1}. Note the factor w – 1 occurs twice; hence, 1 is called a double root or a root of multiplicity two.

  7. (2r + 7) (3r – 8) = 0
  8. Apply the zero factor property immediately.

  9. 4y2 + 2y = 0
  10. 2p (p – 3) = 5p2 – 7p
  11. Distribute and write the equation in standard form first. Then factor.

  12. Multiply both sides by the LCD to clear fractions. Then write the equation in standard form and solve. Check for extraneous solutions.

Follow the procedure employed in part (g): t2 – 9 = (t + 3) (t – 3) = LCD.

We must check for extraneous solutions.

t = 3 is not a solution. The solution set is · t = 3 is an extraneous root. There is only one solution.

We can check our possible solutions by the calculator using the method we employed for solved problem 4.4(e) as shown below for solved problem 6.1 (h):

Since the result for either side was undefined (both sides in this case), 3 is not a solution. It was extraneous. (3 would have also been extraneous if the results for the two sides would have simply been different numbers.)

Practice solving quadratic equations using the factor method by working supplementary problem 6.1.

6.2 Use the square root method to solve the following.

  1. x2 – 49 = 0
  2. Solve for x2 and extract roots.

    The solution set is {±7}. The results may be checked mentally.

  3. y2 + 20 = 0
  4. 5t2 + 1 = 37
  5. (2w – 3)2 = 7
  6. (5x + 1)2 + 13 = 0

See supplementary problem 6.2 for similar exercises.

6.3 Determine the number k such that if k2 is added to the given expression, the result is a perfect square trinomial. Write the result as the square of a binomial.

  1. x2 + 8x
  2. x2 – 8x
  3. t2 + 20t
  4. w2 – 11w

See supplementary problem 6.3.

6.4 Solve by completing the square.

  1. x2 + 10x – 3 = 0
  2. t2 – 7t + 2 = 0
  3. 56y2 – 3y=4

The coefficient of y2 is not one, o begin by

6.5 Use the quadratic formula to solve the following.

  1. x2 – 3x + 1 = 0
  2. The equation is in the appropriate form. Therefore, the coefficients can readily be identified: a = 1, b –3 and c = 1. Substitute the values into the formula and simplify.

  3. 3x2 + 4x – 5 = 0
  4. The equation is in the appropriate form. Therefore, the coefficients can readily be identified; a – 3, b –4 and c = –5. Substitute the values into the formula and simplify.

  5. 3x2 + 4x – 5 = 0
  6. The equation is in the appropriate form. Therefore, the coefficients can readily be identified; a = 3, b = 4 and c =– 5. Substitute the values into the formula and simplify

  7. 3t2 = t – 2
  8. We must first write the equation in the standard form 3t2t + 2 = 0. Now a = 3, b = –1 and c = 2. Substitute into the formula and simplify.

  9. w2 + 3 = –8w

Proceed as in part (c). The standard form is w2 + 8w + 3 = 0. a = 1, b = 8 and c = 3.

Work supplementary problem 6.5 for practice.

6.6 Use your calculator to solve the following. Verify the intermediate values.

  1. 2.36x2 + 4.02x + 1.18 = 0
  2. Use the quadratic formula; a = 2.36, b = 4.02, and c = 1.18.

    The solution set accurate to two decimal places is {–0.0.38,–1.33}.
  3. 1.18x2 – 2.35x+4.03=0

The solution set is {1.00 ± 1.56 i

You were asked to verify the intermediate values in the above problems. Refer to your owner's manual if the values obtained by you were different. Normally it is unnecessary to record the intermediate values in the calculations. They were included for your reference.
The given equations contained coefficients accurate to two decimal places. The answers given are approximate because of round-off, and should be expressed accurate to two decimal places also. In order to accomplish two decimal place accuracy, the intermediate values are given to four decimal place accuracy.

Try supplementary problem 6.6 for additional drill.

6.7 Solve the following. Check for extraneous roots.

  1. Apply Property 1 and solve the resulting equation.

    The solution set is {27}.

  2. Apply Property 1 and solve the resulting equation.

    The solution set is { } = Ø. There is no solution.

  3. Apply Property 1 and solve.

    The solution set is {12}.

  4. Recall that the one-half power of an expression represents the square root of the expression. Apply Property 1 and solve.

    The solution set is {59}.

  5. Isolate the radical term, then apply Property 1 and solve.

  6. Solution 1: Proceed as before.

    Radicals remain, what now? Try a different approach.

    Solution 2:

    Now apply Property 1 and solve.

    The solution set is {5}.

  7. Isolate the radical term, then square and solve.

    t = –4 is an extraneous root. The solution set is {0}.

  8. Apply Property 1 twice, isolating the radical each time.

    w = 7 is an extraneous root. The solution set is {2}.

  9. Apply Property 1 twice.

    x = – 2 x = 1 and are extraneous roots. The solution set is {} = Ø. There is no solution.

    Reflect on the procedure used in the above problems. Property 1 was applied in each case after a radical term was isolated. The result was then an equation we could solve. The squaring step sometimes introduces extraneous solutions. We must check all potential solutions in the original equation prior to specifying the solution set.

    Improve your skills by working supplementary problem 6.7.

    6.8 Solve. Remember to check for extraneous roots.

    1. Apply Property 2. Cubing undoes cube root.

      The solution set is {64}.

    2. Proceed as in part (a).

      The solution set is {12}.

    3. The one-fourth power represents the fourth root of the expression. Apply Property 2.

      y = – 4 is an extraneous solution. There is no solution. The solution set is { } =Ø.

    4. Isolate the radical term first.

      The solution set is {84}.

    5. The variable t will be isolated if we raise both sides to the –3 power.

    6. p = 4 is an extraneous root. There is no solution. The solution set is { } =Ø.

    Isolate the radical term first.

    The solution set is {–3}.

    See supplementary problem 6.8 for similar problems.

    6.9 Solve. Check for extraneous roots when necessary.

    1. y4 – 9y2 + 20 = 0
    2. Since y4 is the square of y2, let u = y2 so that u2 = y4. Now substitute to obtain

    3. t6 + 15 = 8t3
    4. t6 = (t3)2 so let u = t3, so that u2 = t6. Now substitute and rearrange.

    5. A check is mandatory this time.

      Both potential solutions are extraneous. There is no solution. The solution set is { } =Ø

    6. 2w4 + 5w2 – 12 = 0
    7. A check is required.

      The solution set is {–8, 64}.

    8. x_2x_1 – 20 = 0
    9. A check is necessary.

    First clear the fraction, then apply the above process.

    Practice solving quadratic form equations by working supplementary problem 6.9.

    6.10 Solve. State the answer in complete sentence form. A calculator is helpful in many instances.

    1. The distance d in feet that an object falls in t seconds is approximated by the formula d = 16t2. How many seconds elapse if an object falls 2,700 feet?
    2. d = 16t2 so solve 2,700 = 16t2 for t.

      Disregard the negative solution, as it is meaningless.

      Approximately 13 seconds elapse as an object falls 2,700 feet.

    3. The area of a circle is given by the formula A = πr2. Find the radius r of a circular region with area 1, 385 square feet.
    4. A = πr2 so solve 1,385 = πr2 for r. Divide by π, then extract square roots.

      The negative solution is meaningless.

      The radius of the circular region is approximately 21 feet.

    5. The volume V of a cylindrical container is determined by V == π r2h, where r is the radius and h is the height of the cylinder. Five cubic feet of planting mix are used to fill a cylindrical planter 18 inches in height. What is the radius of the container?
    6. Isolate r2 and extract square roots. The height must be expressed in feet since the volume is given in cubic feet. Eighteen inches is = 1.5 feet. V = πr2h or 5 = π r2 (1.5) so

      The negative solution is not applicable.

      The container has a radius of approximately 1.03 feet.

    7. The Pythagorean Theorem states that the sum of the squares of the lengths of the legs, a and b, in a right triangle is equal to the square of the length of the hypotenuse c. That is, a2 + b2 = c2. Find the length of a leg of a right triangle with hypotenuse c = 26 cm and leg a = 24 cm in length.

       

      Suppose b is the length in cm of the unknown leg. Then

      Reject the negative root since lengths are positive.

      The unknown leg is 10 cm in length.

    8. The converse of the Pythagorean Theorem can be employed to construct a right angle. The converse states that if a2 + b2 = c2, where a, b, and c are the lengths of the sides, then the triangle is a right triangle. A right angle is formed by the two shorter sides of the triangle. Find the appropriate length of the longest side of a triangle whose shorter sides are 9 and 12 feet in length in order for the shorter sides to form a right angle.
    9. We must determine the length c when a = 9 and b= 12. We know a2 + b2 = c2, so substitute and solve for c.

      Disregard the negative root.

      The shorter sides form a right angle if the longest side is 15 feet in length.

    10. The height h in feet of a baseball above the ground t seconds after it is thrown upward is given by h= –16t2 + 80t + 5. How long after it is thrown does the ball strike the ground?
    11. The ball strikes the ground when the height h = 0. Use the quadratic formula to solve 0 = –16t2 + 80t + 5 or 16t2 – 80t – 5 – 0 for t.

      The other root is meaningless since it is negative.

      The expression differs from 0 slightly because of round-off. The ball strikes the ground approximately 5 seconds after it is thrown upward.

    12. Find the dimensions of a rectangle whose area is 1,568 m2 if the length is twice its width. The area of a rectangle is A = lw.
    13. Let w be the width of the rectangle in meters. The length is then 2w meters. Therefore, solve

      Disregard the negative solution. If w = 28, l = 2w = 2(28) = 56.

      The dimensions are l = 56 m and w = 28 m.

    14. An airplane requires 4 hrs, 45 min to complete a round trip of 1,100 miles each way. The plane had an 80 mph tailwind going and a 70 mph headwind on the return trip. Find the speed of the plane in still air.

    Let s be the speed in mph of the plane in still air. The rate the plane travels with an 80 mph tailwind is then s – 80 mph. The rate the plane travels with a 70 mph headwind is s – 70 mph. This is a distance, rate, time problem. The quantities are related by the formula d = rt. We need the equivalent form t = d/r. We summarize the above information in the table below.

    The equation which relates the known and unknown quantities can be formed by relating the times involved.

        Total time = time going + time returning

    The times must be expressed in the same units in the equation. We choose hours since we desire the speed of the plane in miles per hour.

    Use the quadratic formula.

    s is negative if we evaluate (2,152.5 – 2,312.5)/9.5. Speed is not negative, so disregard the negative root.

    The speed of the plane in still air is 470 mph.

    Refer to supplementary problem 6.10 for similar applications.

    6.11 Graph the following.

    1. y = x2
    2. We begin by choosing arbitrary values for x and determine the corresponding y values. Make a table of values. Plot the points and connect them with a smooth curve.

      Second-Degree Equations and Inequalities

    3. y = x2 + 2
    4. Proceed as in part (a).

      Second-Degree Equations and Inequalities

    5. y= –x2 + 3 = 3 –x2
    6. Proceed as in part (a).

      Second-Degree Equations and Inequalities

    7. y= –2x2

    Proceed as in part (a).

    Second-Degree Equations and Inequalities

      Observe that the parabola opens upward when a, the coefficient of x2, is positive. The parabola opens downward when a is negative. We can therefore predict the direction the parabola opens merely by observing the sign of the coefficient of the x2 term.

    6.12 Find the maximum or minimum point and graph the expression.

    1. y = x2 – 6x + 5
    2. Make the expression on the right involve the square of a binomial.

      Second-Degree Equations and Inequalities

      The parabola opens upward since a = 1 > 0. The smallest value the right side can have occurs when x = 3. The value of x which makes the expression in parentheses zero is the x-coordinate of the minimum point. Choose x's equidistant from 3 and evaluate to find additional points on the curve. Plot the points and draw a smooth curve through them.

      Second-Degree Equations and Inequalities

    3. y= –x2 – 4x + 1

    Second-Degree Equations and Inequalities

    The parabola opens downward since a= –1 < 0. The largest size the right side can have occurs when x = –2. The value of x which makes the expression in parentheses zero is the x-coordinate of the maximum point. Choose x's equidistant from –2 and evaluate to find additional points on the parabola. Plot the points and draw a smooth curve through them.

    Second-Degree Equations and Inequalities

    6.13 Graph the following. Utilize the procedure stated above.

    1. y = x2 – 4x
      1. The parabola opens upward since a = 1 > 0.
      2. The vertex occurs at Second-Degree Equations and Inequalities
      3. The y-coordinate of the vertex is y = 22 = 4(2) = 4 – 8 = –4.
      4. If x = 0, y = 02 – 4(0) = 0 = 0 = 0. This is the y-intercept.
      5. Ify = 0, 0 = x2 – 4x or x2 = 4x = 0. Hence x(x – 4) = 0.
      6. Therefore, x = 0 and x = 4 are the x-intercepts.

      7. We record the ordered pairs found in a table and find a couple of additional points to graph. Plot the points found and draw a smooth curve through them to complete the graph.
    2. Second-Degree Equations and Inequalities

    3. y=–2x2 – 7x – 3
      1. The parabola opens downward since a = –2 < 0.
      2. The vertex occurs at
      3. If x = 0, y= –2(0)2 – 7(0) – 3 = –3. This is the y-intercept.
      4. Record the ordered pairs found in a table. Find additional points, plot all points, and draw the curve.
    4. y = –2x2 + x – 2
    1. The parabola opens downward since a= –2 < 0.
    2. The vertex is at
    3. If x = 0, y = 2. This is the y-intercept.
    4. There are no x-intercepts since the maximum point has y-coordinate.
    5. Find several more additional points to plot and graph the curve.

6.14 Graph the following.

  1. x = y2 – 3
    1. The parabola opens to the right since a = 1 > 0.
    2. The vertex occurs at
    3. The x-coordinate of the vertex is x = 02 – 3 = –3.
    4. The x-intercept is at the vertex.
    5. Record the values in a table and plot the points. Find about two more points and draw a smooth curve through them.
  2. x = –y2 + 5y + 6
    1. The parabola opens to the right since a = 1 < 0.
    2. The vertex occurs at y = –b/(2a) = –5/(2(–1)) = –5/(–2) = 5/2.
    3. The x-coordinate of the vertex is
    4. If x = 0, 0 = –y2 – 5y + 6 = –(y2 – 5y – 6) = –(y + 1) (y =) and y= –1 or y = 6. These are the y-intercepts.
    5. If y = 0, then x= –02 + 5(0) + 6 = 6 is the x-intercept.
    6. Make a table of the ordered pairs found above. Pick arbitrary values for y to find additional ordered pairs on the curve. Draw the curve.
  3. x= 2y2y + 2
    1. The parabola opens to the right since a = 2 > 0.
    2. The vertex occurs at y= –b/(2a) – (–1)/ (2(2)) = 1/4.
    3. There are no y-intercepts since the vertex lies to the right of the y-axis.
    4. If y = 0, then x = 2(0)2 – 0 + 2 = 2 is the x-intercept.
    5. Ruecord the known ordered pairs in a table. Find additional ordered pairs to plot and draw the graph.

Work supplementary problem 6.11 to practice graphing parabolas.

6.15   Graph the following on a graphing calculator.

  1. y = x2 – 8x + 11
  2. The parabola opens upward since a = 1 > 0. The vertex occurs at x = –b/(2a) = –(–8)/(2(1)) = 8/2 = 4. Try an x–range of [–1, 9]. If x = 4, y = –5 and the y–intercept is 11 so set the y–range to [–6, 12]. Now enter the equation and graph it.

     Second-Degree Equations and Inequalities

  3. y = –x2 – 6x – 5
  4. The parabola opens downward since a = –1 < 0. The vertex occurs at x = –b/ (2a) = – (–6) / (2(–1)) = 6/ (–2) =–3. Set the x–range to [–8, 2]. If x = –3, y = 4 and the y–intercept is –5 so set the y–range to [–10, 5]. Enter the expression and graph it.

     Second-Degree Equations and Inequalities

  5. y = –0.76 (x – 4.38)2 + 7.89
The parabola opens downward since a = –0.76 < 0. The vertex occurs at x = 4.38 since the maximum point occurs at the value of x which makes the expression in parentheses zero. The y–value is 7.89 when x = 4.38. Set the x–range to [–2, 10] and the y–range to [–10, 10]. Graph the expression.

 Second-Degree Equations and Inequalities

6.16   Graph the following on a graphing calculator. Employ the trace feature to approximate the zeros (roots) and vertex of each. Express answers to three decimal place accuracy.

  1. y = x2 – 4.8x + 2.06
  2. Graph the expression. Try to fill the graphing window with the part of the curve that includes the zeros and the vertex. The zoom feature can be utilized to magnify the relevant portion of the curve. Display the y–coordinate and instruct the calculator to trace the curve until y ≈ 0 to determine the first zero. Display the x–coordinate and record it. We obtain x ≈ 0.468 for the first zero. Display the y–coordinate and continue the trace to obtain the vertex. Record the y–coordinate, then display the x–coordinate and record it. The vertex is approximately (2.426, –3.699). Display the y–coordinate and continue the trace to obtain the other zero. Display the x–coordinate and record it. It is x ≈ 4.298.

  3. y =–1.2x2 – 7.92x – 8.368
  4. Proceed as in part (a). The first zero obtained is x≈ –5.279. The vertex is approximately (–3.300, 4.700). The second zero is x ≈ –1.321.

  5. y = 0.77x2 – 6.70x + 16.15

Proceed as in part (a). There are no real zeros. The graph does not intersect the x–axis. The vertex is approximately (4.351, 1.575).

Refer to supplementary problem 6.12 for similar problems.

6.17   Solve the following. Make a sign diagram in each case. Express the solution set in interval notation.

  1. x2 – 3x – 10 > 0
  2. We must find the values of x for which the expression is positive. Employ the procedure stated on p. 175.

    1. x2 + 3x – 10 > 0
    2. Solve x2 + 3x – 10 = 0. x2 + 3x – 10 = (x + 5) (x – 2) = 0 if x = –5 or x = 2. The critical values are –5 and 2.
    3.  Second–Degree Equations and Inequalities
    4. The solution set is  Second–Degree Equations and Inequalities.
  3. 3x2 + x ≤ 10
    1. 3x2 + x = 10 ≤ 0
    2.  Second–Degree Equations and Inequalities
    3.  Second–Degree Equations and Inequalities
    4. The endpoints of the interval in the solution set are included since the critical values satisfy the equality relation. The solution set is .
  4. 2x2 + x ≥ 3
    1. 2x2 + x – 3 ≥ 0
    2.  Second–Degree Equations and Inequalities
    3. The endpoints of the intervals in the solution set are included since the critical values satisfy the equalityrelation. The solution set is .

6.18 Solve the following. Make a sign diagram in each case. Express the solution set in interval notation.

    1. The critical values are the values for which the expression is zero or undefined. The expression is zero if the numerator is zero or if x = –3. The expression is undefined if the denominator is zero or if x = 5. The critical values are –3 and 5.
    2. The solution set is (–3, 5).
    1. The numerator is zero if t = 0. The denominator is zero if t = . The critical values are 0 and
    2. .
    3. The solution set is (–∞, 0] (, ∞). Note that t = . excluded since the expression is undefined at t = .
    1. We must first rewrite the expression with one side of the inequality equal to zero.

    2. The numerator is zero if 11 – 2s = 0 or if s = . The denominator is zero if s or if s = 2. The critical values are and 2.
    3. The solution set is (–∞, 2] [, ∞). It can be verified that values of s in the indicated intervals satisfy

See supplementary problem 6.13 to practice solving quadratic and rational inequalities

6.19 (x – 3)3 (x – 2)5 (x – 1)4 > 0

The critical values are 3, –2, and 1. They break the number line into four intervals: (– ∞, –2), (–2, 1), (1, 3), and (3, ∞). We must determine the solution behavior of each interval. We use the test value of zero to start the process:
x = 0 gives (–)3 (–)5 (–)4 = (–) which is not greater than zero as the inequality

 

requires.
Therefore, the interval containing zero, (–2, 1), is excluded from the solution. See Figure 6.23 below.

The solution set is (– ∞, –2) (3, ∞).

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