4x^4
_____
10x^5
Or simply 4x^4/10x^5
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Step-by-step explanation:
Hey,
(x+2)^2 =(x-5)^2+7,how do I solve for x?
At first here we have to use the formulas (a+b)^2 and (a-b)^2 for solving (x+2)^2 and (x-5)^2 …we know (a+b)^2= a^2+2ab+b^2 and
(a-b)^2= a^2–2ab+b^2
So using these formulas solve for (x+2)^2 and (x-5)^2
(x+2)^2= x^2+4x+4
(x-5)^2=x^2–10x+25
So solve the equation
(x+2)^2=(x-5)^2+7
=>x^2+4x+4=x^2–10x+25+7
=>4x+4=32–10x
=>4x+10x=32–4
=>14x=28
=>x=28/14
=>x=2
If you want to check if your answer is correct or not then you can put the value of x in the equation.
So putting the value of x in the equation we get
(2+2)^2=(2–5)^2+7
=>4^2=(—3)^2+7
=>16=16
Thus,our value of x is correct
So the value of x =2
I hope you may get some help..
Thank you
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