Algebra: Quadratic Formula and Word Problems Review

Algebra Review

Quadratic Concepts

  • Quadratic Formula: The quadratic formula is used to find the roots of a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0. It is given by:   x=racbext±extD2a,x = rac{-b ext{±} ext{√}D}{2a}, where D=b24acD = b^2 - 4ac is the discriminant.

Determining the Nature of Roots Using the Discriminant
  • The discriminant can be used to determine the number of real roots:   - If D > 0: There are 2 distinct real roots.   - If D = 0: There is 1 real repeated root (rational).   - If D < 0: No real roots (complex).   - Roots are rational if they can be expressed as a fraction of integers, and irrational if not.

Exercises
  1. Use the discriminant to determine the number of roots:    - A. x2+12x+36x^2 + 12x + 36
           - Discriminant: D=1224(1)(36)=0D = 12^2 - 4(1)(36) = 0
           - 1 Rational Root.    - B. x28x+10x^2 - 8x + 10
           - Discriminant: D=(8)24(1)(10)=6440=24D = (-8)^2 - 4(1)(10) = 64 - 40 = 24
           - 2 Rational Roots.    - C. x26x+20x^2 - 6x + 20
           - Discriminant: D=(6)24(1)(20)=3680=44D = (-6)^2 - 4(1)(20) = 36 - 80 = -44
           - No Real Roots.    - D. 6x2+19x+106x^2 + 19x + 10
           - Discriminant: D=1924(6)(10)=361240=121D = 19^2 - 4(6)(10) = 361 - 240 = 121
           - 2 Rational Roots.

  2. Use the quadratic formula to find the roots:    - A. x2+10x+4=0x^2 + 10x + 4 = 0
           - x=rac10ext±ext(1004)2=rac10ext±ext(96)2=5ext±4ext(6)x = rac{-10 ext{±} ext{√}(100 - 4)}{2} = rac{-10 ext{±} ext{√}(96)}{2} = -5 ext{±} 4 ext{√}(6).    - B. 2x2+4x4=02x^2 + 4x - 4 = 0
           - x=rac4ext±ext(16+32)4=rac4ext±84=1extor3x = rac{-4 ext{±} ext{√}(16 + 32)}{4} = rac{-4 ext{±} 8}{4} = 1 ext{ or } -3.    - C. x212x+8=0x^2 - 12x + 8 = 0
           - x=rac12ext±ext(14432)2=6ext±ext(112)=6ext±4ext(7)x = rac{12 ext{±} ext{√}(144 - 32)}{2} = 6 ext{±} ext{√}(112) = 6 ext{±} 4 ext{√}(7).

Word Problems
  1. Rectangle Problem:    - Let width = xx, length = x+3x + 3.    - Area = x(x+3)=28x(x + 3) = 28
           x2+3x28=0x^2 + 3x - 28 = 0
           - Factorization gives: (x+7)(x4)=0(x + 7)(x - 4) = 0
           - Width = 4, Length = 7.

  2. Triangle Base and Height:    - Base = 3x3x, Height = 4x4x.    - Area = rac12imesbaseimesheight=54rac{1}{2} imes base imes height = 54
           - rac12imes3ximes4x=54rac{1}{2} imes 3x imes 4x = 54        - 6x2=546x^2 = 54
           -        x2=9ightarrowx=3x^2 = 9 ightarrow x = 3
           - Base = 9, Height = 12.

  3. Jack’s Pan Problem:    - Let size of square to cut = xx.    - Area = 120120
           - Bottom dimensions after cuts: 162x16 - 2x and 142x14 - 2x.        - (162x)(142x)=120(16 - 2x)(14 - 2x) = 120
           - Solve to find cut size: x=2x = 2.

Quadratic-Linear Systems of Equations

  • A quadratic-linear pair consists of a quadratic equation and a linear equation. These can be solved either graphically or algebraically.

Exercises:
  1. Solve using a calculator:
       - y=x26x+5y = x^2 - 6x + 5 and y=2x10y = 2x - 10.
       - Points of intersection: (3,4)(3, -4) and (5,0)(5, 0).

  2. Algebraic solution for:    - y=x2+5y = x^2 + 5 and 3x+2y=103x + 2y = 10    - Substitute: 3x+2(x2+5)=103x + 2(x^2 + 5) = 10    - Quadratic: 2x2+3x6=02x^2 + 3x - 6 = 0
       - Roots: x=3x = 3 or x=1x = -1 (found).
       - Find corresponding yy for each x.

  3. For:    - 4xy=34x - y = 3 and y=2x2+12x9y = -2x^2 + 12x - 9    - Solve by substitution and factorization to obtain roots etc. Resulting checks give the solutions.

Quadratic Problems
  1. Reciprocal Problem:    - Let number = xx. Therefore: x+rac1x=10x + rac{1}{x} = 10
             Solve to find the numbers.

  2. Equation of a Quadratic: Identify the equation based on the graph provided.       

  3. Turning Point Calculation for y=3x224x+7y = 3x^2 - 24x + 7:    - Vertex formula: x=racb2a=rac246=4x = - rac{b}{2a} = rac{24}{6} = 4
       - Substituting back gives y=41y = -41.    - Thus, vertex = (4,41)(4, -41).

  4. Axis of Symmetry: For roots x=4x = 4 and x=14x = 14, the axis of symmetry is x=rac4+142=9x = rac{4 + 14}{2} = 9.

Further Problems and Applications
  1. Solve: y=10x2+x3y = 10x^2 + x - 3 for roots and other properties such as vertex, intercepts, etc.

  2. Given quadratic y=x24x12y = x^2 - 4x - 12, determine all relevant features including axis of symmetry, vertex, and roots.

  3. Given the conditions and a root of 1 for the quadratic, determine k and second root.

  4. Similar for another quadratic involving finding k.

  5. Graphical Method: Solve the given system of equations graphically, confirming the intersections of curves.

  6. Find quadratic equations from set roots.

  7. Find dimensions of geometric figures based on provided conditions.

  8. Discriminant analysis for the roots of quadratics and describability.

  9. Logic puzzle involving integers based on given conditions.

  10. Geometry of triangles with specific dimensions based on side length relationships.

  11. Rectangle dimensions based on perimeter and area.

  12. Right triangle dimensions according to the ratio of sides and hypotenuse.