Solving the Quadratic: Tackling (x-5)² - 5 = -12 + 2x

Quadratic Equations with Expanded Form

Solve the following equation:

(x5)25=12+2x (x-5)^2-5=-12+2x

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find X
00:03 Use the abbreviated multiplication formulas
00:16 Substitute appropriate values according to the given data and expand the brackets
00:37 Substitute in our equation
00:57 Arrange the equation so that one side equals 0
01:12 Collect like terms
01:37 Examine the coefficients
01:45 Use the quadratic formula
02:07 Substitute appropriate values and solve
02:26 Calculate the square and multiplications
02:59 Calculate the square root of 16
03:06 These are the 2 possible solutions (addition, subtraction)
03:28 And this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve the following equation:

(x5)25=12+2x (x-5)^2-5=-12+2x

2

Step-by-step solution

To solve the equation (x5)25=12+2x(x-5)^2 - 5 = -12 + 2x, follow these steps:

  • Step 1: Expand the square on the left side of the equation:
    (x5)2=x210x+25(x-5)^2 = x^2 - 10x + 25
  • Step 2: Substitute this back into the equation:
    x210x+255=12+2xx^2 - 10x + 25 - 5 = -12 + 2x
  • Step 3: Simplify the equation:
    x210x+20=12+2xx^2 - 10x + 20 = -12 + 2x
  • Step 4: Rearrange the equation by moving all terms to one side:
    x210x+202x+12=0x^2 - 10x + 20 - 2x + 12 = 0
    This simplifies to x212x+32=0x^2 - 12x + 32 = 0.
  • Step 5: Use the Quadratic Formula, where a=1a = 1, b=12b = -12, and c=32c = 32:
    x=(12)±(12)24×1×322×1x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \times 1 \times 32}}{2 \times 1}
  • Step 6: Calculate the discriminant and simplify:
    x=12±1441282x = \frac{12 \pm \sqrt{144 - 128}}{2}
    x=12±162x = \frac{12 \pm \sqrt{16}}{2}
    x=12±42x = \frac{12 \pm 4}{2}
  • Step 7: Solve for the two potential values of xx:
    x1=12+42=8x_1 = \frac{12 + 4}{2} = 8
    x2=1242=4x_2 = \frac{12 - 4}{2} = 4

Thus, the solutions to the equation are x1=8x_1 = 8 and x2=4x_2 = 4.

Therefore, the correct answer is x1=8,x2=4x_1 = 8, x_2 = 4, which corresponds to choice 1.

3

Final Answer

x1=8,x2=4 x_1=8,x_2=4

Key Points to Remember

Essential concepts to master this topic
  • Expansion Rule: Apply (ab)2=a22ab+b2 (a-b)^2 = a^2 - 2ab + b^2 formula correctly
  • Standard Form: Move all terms to one side: x212x+32=0 x^2 - 12x + 32 = 0
  • Verification: Check both solutions: (85)25=12+2(8) (8-5)^2 - 5 = -12 + 2(8) gives 4=4 4 = 4

Common Mistakes

Avoid these frequent errors
  • Expanding (x-5)² incorrectly
    Don't expand (x5)2 (x-5)^2 as x225 x^2 - 25 = missing the middle term! This gives wrong coefficients in your quadratic equation. Always use the formula: (x5)2=x210x+25 (x-5)^2 = x^2 - 10x + 25 .

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 can't I just expand (x-5)² as x² - 25?

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That's a common mistake! (x5)2 (x-5)^2 means (x-5)(x-5), not x² - 5². You need the middle term: x25x5x+25=x210x+25 x^2 - 5x - 5x + 25 = x^2 - 10x + 25 .

Should I expand the left side first or move terms around?

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Always expand first, then simplify, then move all terms to one side. This prevents errors and keeps your work organized step by step.

How do I know which method to use - factoring or quadratic formula?

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Try factoring first! For x212x+32=0 x^2 - 12x + 32 = 0 , look for two numbers that multiply to 32 and add to -12. If factoring seems difficult, the quadratic formula always works.

Why do I get two different answers?

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Quadratic equations typically have two solutions because a parabola crosses the x-axis at two points. Both x = 8 and x = 4 are correct solutions to this equation!

What if my discriminant is negative?

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A negative discriminant means no real solutions exist. In this problem, 144128=16>0 144 - 128 = 16 > 0 , so we have two real solutions.

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