anonymous
  • anonymous
I really need some help with these questions... 1. 2x^2-5x+2=(x-3)(x-2)+3x 2. x(2x-5)=12 3. x(x+7)=14 4. x+1-2(square root of x+4=0) 5. |2x-5|4-|x-3| 6. 2x-3=x^3-5 7. (x+1)^-1=x^-1-x
Precalculus
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
For #1: FOIL the right-hand side and combine like terms. Then you will see the rest as to what you can add or subtract to get your solution. For #2: Distribute the left-hand side first. Then, subtract 12 from both sides of the equation to get something like: \[ax^2 + bx + c = 0\]. Factor the polynomial equation to get your solution for \[x \] For #3: See #2. For #4-#7: can't see the problems properly. Rewrite so that I can get a better picture.
anonymous
  • anonymous
4. \[x+1-2\sqrt{x+4=0}\] 5. \[\left| 2x-5 \right|4-\left| x-3 \right|\] 6. \[2x-3=x ^{3}-5\] 7. \[x+1^{-1}=x ^{-1}-x\]
anonymous
  • anonymous
#4: are we solving for values of x? #5: I need to think about it a little more. #6: rearrange to get \[x^3-0x^2-2x-2 = 0\] you may have to do long division. #7: I interpret it as follows:\[x + 1 = (1/x) - x\] since the inverse of 1 is 1, and, x to the negative power is the inverse as shown. then multiply both sides by x to get \[x^2 + x = 1 - x^2 \] add an\[x^2 \] to both sides to get \[2x^2 + x = 1\] then subtract a 1 on both sides to get \[2x^2 + x -1 = 0\] solve for x to get solution

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