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A.8(D)·Algebra II·Quadratic & Square Root FunctionsReadiness

Solving Quadratic Equations

Solve quadratic equations using factoring, tables, and graphs.

Students solve quadratics by factoring and substitution, then use tables and graphs to locate roots that are not listed exactly.

Lesson plan and worked examples by TEKSForge · not a TEA-issued lesson. Videos by independent educators.

Archived course: TEKS code and readiness label from TEA's 2013 Algebra II key. Algebra II is not one of Texas's current required STAAR EOC assessments, and this lesson does not change the Biology and Algebra I Study Plans.

Practice the released item Ask the tutorAll Algebra II released practice50 minutes

Learn the concept

Make a zero product

Move every term to one side, factor, and set each factor equal to zero. For example, a factor 3x−23x-2 produces x=23x=\frac23, not −23-\frac23.

Check in the original equation

Substitution can verify a proposed answer or eliminate a choice. Use the original equation so an earlier rearrangement error does not become part of the check.

Locate roots between rows

Quadratic functions are continuous. Opposite-sign outputs at two inputs guarantee a zero between them. Refine a calculator table or inspect the graph to narrow that interval.

Worked example

Solve 6x2+x=26x^2+x=2 and verify the positive solution.

  1. Write 6x2+x−2=06x^2+x-2=0.
  2. Factor: (3x+2)(2x−1)=0(3x+2)(2x-1)=0.
  3. Solve each factor: x=−23x=-\frac23 or x=12x=\frac12.
  4. Check the positive solution: 6(12)2+12=32+12=26\left(\frac12\right)^2+\frac12=\frac32+\frac12=2.

x=−23x=-\frac23 or x=12x=\frac12.

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Texas public-school, Texas educator-created, and TEKS-aligned lessons, hosted by their creators. Each card names its source.

Texas public school video

Algebra 2 Solving Quadratic Equations

UT IPSI · Kid2Kid / Texas Education Agency

Use algebraic and graphical methods to solve and verify quadratic equations.

Watch on YouTube

Try it yourself

TEKSForge practice · unscored

  1. 1. Solve 8x2−2x=38x^2-2x=3.

    Show answer and reasoning

    x=34x=\frac34 or x=−12x=-\frac12

    8x2−2x−3=(4x−3)(2x+1)=08x^2-2x-3=(4x-3)(2x+1)=0.

  2. 2. For q(x)=x2−3x−1q(x)=x^2-3x-1, find consecutive integers between which each zero lies.

    Show answer and reasoning

    −1<x<0-1<x<0 and 3<x<43<x<4

    q(−1)=3q(-1)=3, q(0)=−1q(0)=-1, q(3)=−1q(3)=-1, and q(4)=3q(4)=3. Both intervals have sign changes.

Released STAAR practice

Texas Education Agency · archived 2013 released EOC items · scored with TEA's answer key; step-by-step explanations written by TEKSForge because this release has no rationales. Original graphs, tables, and diagrams are shown where the released item includes one.

Archived released STAAR Algebra II EOC · 2013 · Item 18 · 1 point

Which of the following is a solution to 10x2−x=310x^2 - x = 3?

  1. −15-\frac{1}{5}
  2. 35\frac{3}{5}
  3. −32-\frac{3}{2}
  4. 12\frac{1}{2}
Show STAAR answer and explanation

35\frac{3}{5}

Rewrite as 10x2−x−3=010x^2-x-3=0 and factor: (5x−3)(2x+1)=0(5x-3)(2x+1)=0. The solutions are x=35x=\frac35 and x=−12x=-\frac12; only 35\frac35 is a choice. Check: 10(925)−35=185−35=310\left(\frac{9}{25}\right)-\frac35=\frac{18}{5}-\frac35=3.

−15-\frac{1}{5}: 10(125)+15=3510\left(\frac{1}{25}\right)+\frac15=\frac35, not 3.

−32-\frac{3}{2}: 10(94)+32=2410\left(\frac94\right)+\frac32=24, not 3.

12\frac{1}{2}: 10(14)−12=210\left(\frac14\right)-\frac12=2, not 3. The other solution is −12-\frac12, not 12\frac12.

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Archived released STAAR Algebra II EOC · 2013 · Item 27 · 1 point

The table shows values of a quadratic function gg. Based on the table, in which interval can a solution to g(x)=0g(x) = 0 be found?

Values of g
xg(x)
−5−11
−48
−321
−228
−129
024
113
2−4
3−27
  1. −5<x<−4-5 < x < -4
  2. −2<x<−1-2 < x < -1
  3. −1<x<1-1 < x < 1
  4. 2<x<32 < x < 3
Show STAAR answer and explanation

−5<x<−4-5 < x < -4

g(−5)=−11g(-5)=-11 and g(−4)=8g(-4)=8. The output changes from negative to positive, so the graph crosses the x-axis between x=−5x=-5 and x=−4x=-4.

−2<x<−1-2 < x < -1: g(−2)=28g(-2)=28 and g(−1)=29g(-1)=29 are both positive, so there is no sign change.

−1<x<1-1 < x < 1: g(−1)=29g(-1)=29, g(0)=24g(0)=24, and g(1)=13g(1)=13 are all positive.

2<x<32 < x < 3: g(2)=−4g(2)=-4 and g(3)=−27g(3)=-27 are both negative. The other zero lies between 1 and 2.

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Start scored STAAR practice

Learning objectives

By the end of the lesson, students can:

  • Set a quadratic equation equal to zero and factor it.
  • Verify rational solutions in the original equation.
  • Bracket a real zero using a sign change in a table.

Instructional sequence

  1. 1

    Engage

    5 min

    Test a proposed fractional solution in a quadratic equation.

  2. 2

    Direct instruction

    12 min

    Model standard form, factoring, and the zero-product property.

  3. 3

    Guided practice

    13 min

    Solve quadratics with rational roots and verify the results.

  4. 4

    Table and graph check

    13 min

    Locate sign changes and narrow intervals containing zeros.

  5. 5

    Closure

    7 min

    Students distinguish an exact solution from a bracketing interval.

Misconceptions to address

““Set each term equal to zero before factoring.””

Teacher move: The zero-product property applies to a product equal to zero, not a sum.

““A table has no root unless an output is exactly zero.””

Teacher move: A continuous quadratic can cross the axis between listed inputs when outputs change sign.