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How do you solve a quadratic equation, and how do you choose between factoring, square roots, and the formula?

Solve quadratic equations by factoring with the zero-product property, by taking square roots, and by the quadratic formula, use the discriminant to count real roots, and interpret solutions in context.

A Grade 10 Math MCAS answer on solving quadratics by factoring (zero-product property), taking square roots, and the quadratic formula, using the discriminant to count real roots, and discarding solutions that make no sense in context.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. Set to zero, then factor
  3. Taking square roots
  4. The quadratic formula
  5. Choosing a method and interpreting in context
  6. Try this

What this topic is asking

The Algebra category requires solving quadratic equations three ways (the A-REI standards): factoring, taking square roots, and the quadratic formula. The Grade 10 MCAS asks you to choose an efficient method, to count real roots with the discriminant, and to interpret solutions in a context such as projectile height or area. Quadratics appear in selected-response items and reliably in a constructed-response problem, where method credit is awarded for showing setup and steps.

Set to zero, then factor

The zero-product property is the engine of factoring: a product equals zero only when one of its factors is zero. So the equation must be set equal to zero before factoring.

x2+5x=24  ⇒  x2+5x−24=0  ⇒  (x+8)(x−3)=0  ⇒  x=−8 or x=3.x^2 + 5x = 24 \;\Rightarrow\; x^2 + 5x - 24 = 0 \;\Rightarrow\; (x + 8)(x - 3) = 0 \;\Rightarrow\; x = -8 \text{ or } x = 3.

The single most common error is factoring while one side is still nonzero. Writing x(x+5)=24x(x + 5) = 24 tells you nothing useful, because 24 is not zero and the zero-product property does not apply.

Taking square roots

When a quadratic has no linear term, isolate the squared quantity and take the square root of both sides, remembering the ±\pm.

x2=49  ⇒  x=±7,(x−3)2=16  ⇒  x−3=±4  ⇒  x=7 or x=−1.x^2 = 49 \;\Rightarrow\; x = \pm 7, \qquad (x - 3)^2 = 16 \;\Rightarrow\; x - 3 = \pm 4 \;\Rightarrow\; x = 7 \text{ or } x = -1.

Forgetting the negative root halves your answers and is a frequent slip. This method is fastest for any equation already in the form (something)2=^2 = constant.

The quadratic formula

When a quadratic does not factor with integers, the quadratic formula always works, and it is provided on the MCAS reference sheet.

Because the formula is on the sheet, the credit is for substituting correctly and simplifying, especially writing a radical in simplest form (so 72=62\sqrt{72} = 6\sqrt{2}) when exact answers are required. Watch the sign of −4ac-4ac: when cc is negative, −4ac-4ac becomes positive and adds to b2b^2.

Choosing a method and interpreting in context

Match the method to the equation. If it factors, factoring is fastest. If there is no linear term, take square roots. If nothing factors, use the formula. In a contextual quadratic, the solutions are times, lengths, or counts, so a negative root is usually rejected. For a ball with height h(t)=−16t2+48th(t) = -16t^2 + 48t, setting h=0h = 0 gives t(−16t+48)=0t(−16t + 48) = 0, so t=0t = 0 (launch) and t=3t = 3 (landing); a negative time would be discarded.

A useful check: the methods always agree, so a solution found by the formula should satisfy the original equation when substituted back.

Try this

Q1. Solve (x+2)(x−9)=0(x + 2)(x - 9) = 0.

  • Cue. x=−2x = -2 or x=9x = 9.

Q2. How many real roots does x2+x+4=0x^2 + x + 4 = 0 have?

  • Cue. Discriminant 1−16=−15<01 - 16 = -15 < 0, so no real roots.

Exam-style practice questions

Practice questions written in the style of MA DESE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Grade 10 Math MCAS (style)1 marksSelected-response. What are the solutions to x2−2x=24x^2 - 2x = 24? (A) x=6x = 6 and x=−4x = -4 (B) x=−6x = -6 and x=4x = 4 (C) x=12x = 12 and x=2x = 2 (D) x=24x = 24 and x=0x = 0
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The correct answer is (A).

Set to zero first: x2−2x−24=0x^2 - 2x - 24 = 0. Factor: two numbers with product −24-24 and sum −2-2 are −6-6 and 44, so (x−6)(x+4)=0(x - 6)(x + 4) = 0. By the zero-product property x=6x = 6 or x=−4x = -4. Check x=6x = 6: 36−12=2436 - 12 = 24. The common error is factoring before moving 24 to the left, or swapping the signs to get (B).

Grade 10 Math MCAS (style)4 marksConstructed-response. Solve 2x2+6x−5=02x^2 + 6x - 5 = 0 using the quadratic formula. Give the solutions in simplest radical form and show all steps.
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A 4-point constructed-response: credit for the substitution, the discriminant, the simplification, and the radical form.

With a=2a = 2, b=6b = 6, c=−5c = -5: x=−6±62−4(2)(−5)2(2)=−6±36+404=−6±764x = \frac{-6 \pm \sqrt{6^2 - 4(2)(-5)}}{2(2)} = \frac{-6 \pm \sqrt{36 + 40}}{4} = \frac{-6 \pm \sqrt{76}}{4}. Simplify 76=219\sqrt{76} = 2\sqrt{19}: x=−6±2194=−3±192x = \frac{-6 \pm 2\sqrt{19}}{4} = \frac{-3 \pm \sqrt{19}}{2}. A sign error in −4ac-4ac (it is +40+40 because cc is negative), or leaving 76\sqrt{76} unsimplified, costs a point. A rounded decimal when simplest radical form is required also loses credit.

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