How do you solve # (x-6)^2 = 25#?

Answer 1

#x=1# or #x=11#

Rewriting #(x-6)^2=25# as #(x-6)^2-5^2=0#
we see the left side is the difference of squares with factors: #((x-6)+5)*((x-6)-5) = 0#

Either #{: (,(x-6)+5 = 0, ," or ",,(x-6)-5=0), (rArr,x-1=0, , , ,x-11=0), (rArr,x=1, ," or ", ,x=11) :}#

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Answer 2

To solve the equation (x-6)^2 = 25:

  1. Expand the left side of the equation: (x-6)^2 = (x-6)(x-6) = x^2 - 12x + 36.
  2. Set the expanded expression equal to 25: x^2 - 12x + 36 = 25.
  3. Rearrange the equation by subtracting 25 from both sides: x^2 - 12x + 36 - 25 = 0, which simplifies to x^2 - 12x + 11 = 0.
  4. Factor the quadratic equation: (x - 11)(x - 1) = 0.
  5. Set each factor equal to zero and solve for x:
    • x - 11 = 0 => x = 11
    • x - 1 = 0 => x = 1.

So, the solutions to the equation are x = 11 and x = 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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