How do you solve #(x + 3) ( x - 1) = 32#?

Answer 1

#x=-7,5#

Solve:

#(x+3)(x-1)=32#

FOIL the left-hand side. https://tutor.hix.ai

#x^2+2x-3=32#

Move all terms to the left-hand side.

#x^2+2x-3-32=0#
#x^2+2x-35=0#
Find two numbers that when added equal #2# and when multiplied equal #-35#. The numbers #7# and #-5# meet the criteria.
#(x+7)(x-5)+0#
#x+7=0#
#x=-7#
#x-5=0#
#x=5#
#x=-7,5#
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Answer 2
First of all, you do the multiplications, so you get: #x^2+2x-3=32# Then you put the 32 on the first class just like: #x^2+2x-35=0# Then you do #D=2^2-4*(-35) = 144# (Using the method #D = b^2-4ac#) so your one root is #x_1=(b-sqrt(D))/(2a)# and the other one is #x_2=(b+sqrt(D))/(2a)# Do the maths and you get #x_1=-7# , #x_2=5#
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Answer 3

To solve the equation (x + 3) (x - 1) = 32:

  1. Expand the expression: x^2 + 3x - x - 3 = 32.
  2. Combine like terms: x^2 + 2x - 3 = 32.
  3. Rearrange the equation: x^2 + 2x - 3 - 32 = 0.
  4. Combine like terms again: x^2 + 2x - 35 = 0.
  5. Factor the quadratic equation: (x + 7)(x - 5) = 0.
  6. Set each factor equal to zero and solve for x: x + 7 = 0 => x = -7 x - 5 = 0 => x = 5.

So the solutions are x = -7 and x = 5.

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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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