Solve the Quadratic Equation: 2(a-4)² + 3 = 163-16a

Quadratic Equations with Perfect Square Factoring

Find a

2(a−4)2+3=163−16a 2(a-4)^2+3=163-16a

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find A
00:04 Use shortened multiplication formulas to open the parentheses
00:32 Solve the multiplications and squares
00:44 Open parentheses properly, multiply by each factor
01:03 Simplify what we can
01:14 Isolate A
01:40 Extract the root
01:44 When extracting a root, there are always 2 possibilities: negative and positive
01:47 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find a

2(a−4)2+3=163−16a 2(a-4)^2+3=163-16a

2

Step-by-step solution

To solve the given equation 2(a−4)2+3=163−16a2(a-4)^2 + 3 = 163 - 16a, we begin by expanding the expression (a−4)2(a-4)^2.

  • Step 1: Expand (a−4)2(a-4)^2: (a−4)2=a2−8a+16(a-4)^2 = a^2 - 8a + 16.
  • Step 2: Multiply the expanded form by 2: 2(a2−8a+16)=2a2−16a+322(a^2 - 8a + 16) = 2a^2 - 16a + 32.
  • Step 3: Substitute this into the equation and combine like terms: 2a2−16a+32+3=163−16a2a^2 - 16a + 32 + 3 = 163 - 16a. This simplifies to 2a2−16a+35=163−16a2a^2 - 16a + 35 = 163 - 16a.
  • Step 4: Move all terms to one side of the equation: 2a2−16a−16a+35−163=02a^2 -16a - 16a + 35 - 163 = 0 2a2−32a−128=02a^2 - 32a - 128 = 0.
  • Step 5: Simplify the quadratic equation by dividing by 2: a2−16a−64=0a^2 - 16a - 64 = 0.
  • Step 6: Solve the quadratic equation by factoring: a2−16a−64=(a−8)2−64=(a−8)(a+8)=0a^2 - 16a - 64 = (a - 8)^2 - 64 = (a - 8)(a + 8) = 0.

Finding the roots gives a−8=0a - 8 = 0 or a+8=0a + 8 = 0. Thus, a=8a = 8 or a=−8a = -8.

Therefore, the solutions to the equation are ±8\pm 8.

Thus, the correct answer is ±8\pm 8.

3

Final Answer

±8 \pm8

Key Points to Remember

Essential concepts to master this topic
  • Expansion Rule: Always expand (a−4)2(a-4)^2 before combining like terms
  • Technique: Factor a2−16a+64a^2 - 16a + 64 as (a−8)2(a-8)^2 for easier solving
  • Check: Substitute both a=8a = 8 and a=−8a = -8 into original equation ✓

Common Mistakes

Avoid these frequent errors
  • Incorrectly expanding the squared binomial
    Don't expand (a−4)2(a-4)^2 as a2−16a^2 - 16 = missing the middle term! This creates 2a2−8+3=163−16a2a^2 - 8 + 3 = 163 - 16a which gives wrong solutions. Always use the formula (a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2 to get a2−8a+16a^2 - 8a + 16.

Practice Quiz

Test your knowledge with interactive questions

Declares the given expression as a sum

\( (7b-3x)^2 \)

FAQ

Everything you need to know about this question

Why do I get two answers for this quadratic equation?

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Quadratic equations typically have two solutions because when you take the square root of both sides, you get both positive and negative values. In this case, both a=8a = 8 and a=−8a = -8 satisfy the original equation.

How do I expand (a−4)2(a-4)^2 correctly?

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Use the formula (a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2. So (a−4)2=a2−2(a)(4)+42=a2−8a+16(a-4)^2 = a^2 - 2(a)(4) + 4^2 = a^2 - 8a + 16. Don't forget the middle term!

Can I solve this equation without expanding the square?

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While expanding is the standard method, you could rearrange to get 2(a−4)2=160−16a2(a-4)^2 = 160 - 16a, but this approach is usually more complex. Expanding first typically leads to cleaner solutions.

What if I can't factor the quadratic easily?

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If factoring seems difficult, try the quadratic formula: a=−b±b2−4ac2aa = \frac{-b \pm \sqrt{b^2-4ac}}{2a}. For this problem, you'd get the same answers: a=8a = 8 or a=−8a = -8.

How do I verify my answers are correct?

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Substitute each solution back into the original equation. For a=8a = 8: 2(8−4)2+3=2(16)+3=352(8-4)^2 + 3 = 2(16) + 3 = 35 and 163−16(8)=35163 - 16(8) = 35 ✓

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