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klemol [59]
2 years ago
11

4x+5=x(x+8) how do I solve this?

Mathematics
1 answer:
GenaCL600 [577]2 years ago
3 0

Answer:

x=1,-5

Step-by-step explanation:

We\ are\ given\ that,\\4x+5=x(x+8)\\Hence\ by\ expanding\ the\ terms,\\4x+5=x^2+8x\\Hence\ by\ moving\ 4x\ and\ 5 to\ the\ RHS\ ,\\0=x^2+8x-4x-5\\Hence\ by\ simplifying\ the\ RHS,\\0=x^2+4x-5\\Now,\\We\ take\ a\ closer\ look\ at\ the\ RHS,\\x^2+4x-5\ can\ also\ be\ written\ as:\\=x^2+5x-x-5\\Now, By\ factorizing,\\x(x+5)-1(x+5)\\=(x-1)(x+5)\\Hence,\\(x-1)(x+5)=0\\Now,\\To\ nullify\ the\ LHS\ to\ 0,\ either\ the\ term\ (x-1)\ or/and\ (x+5)\ should\ equate\ to\ 0.\\Hence,\\x-1=0\\x=1\ is\ our\ first\ solution\\(x+5)=0\\x=-5\ is\ our\ second\ solution\\Hence,\\x=1,-5

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==========================================================

Further explanation:

Let's go through the answer choices one by one

  • A. This is false because 0 does not come before -5, but instead -5 is listed first. The order -5,-4,-3,-2,-1,0 is correct meaning that -5 \le x \le 0 is the correct order as well.
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  • C. This is false. The student used the correct inequality signs to indicate x is -5 or larger and also 0 or smaller; basically x is between -5 and 0 inclusive of both endpoints. The "or equal to" portions indicate we are keeping the endpoints and not excluding them.
  • D. This is false. Writing -5 \ge x \ge 0 would not make any sense. This is because that compound inequality breaks down into -5 \ge x \ \text{ and } \ x \ge 0. Try to think of a number that is both smaller than -5 AND also larger than 0. It can't be done. No such number exists.
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