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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(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.. 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 you