How do you solve #12 = 2x^2 - 5x#?

Answer 1

#x=-3/2" or "x=4#

#"rearrange the equation into "color(blue)"standard form"#
#rArr2x^2-5x-12=0larrcolor(blue)"in standard form"#
#"factorise using the a-c method"#
#"that is factors of the product "2xx-12=-24# #"which sum to - 5"#
#"the factors required are - 8 and + 3"#
#"'split' the middle term using these factors"#
#rArr2x^2-8x+3x-12=0larrcolor(blue)"factor by grouping"#
#rArrcolor(red)(2x)(x-4)color(red)(+3)(x-4)=0#
#"take out the common factor "(x-4)#
#rArr(x-4)(color(red)(2x+3))=0#
#"equate each factor to zero and solve for x"#
#x-4=0rArrx=4#
#2x+3=0rArrx=-3/2#
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Answer 2

To solve the equation 12 = 2x^2 - 5x:

  1. Rearrange the equation to set it equal to zero: 2x^2 - 5x - 12 = 0.
  2. Factor the quadratic equation: (2x + 3)(x - 4) = 0.
  3. Set each factor equal to zero and solve for x: 2x + 3 = 0, which gives x = -3/2. x - 4 = 0, which gives x = 4.
  4. The solutions to the equation are x = -3/2 and x = 4.
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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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