 Custom Algebra Teaser.. Answer this pls..? - XP Math - Forums Sign Up FREE! | Sign In | Classroom Setup | Common Core Alignment  XP Math - Forums Custom Algebra Teaser.. Answer this pls..? 07-25-2007 #1 Alvin C. Guest   Posts: n/a Custom Algebra Teaser.. Answer this pls..? x^2+a(a-2x)=a^2+x(x-2a)How can I get the answer? My teacher let us answer this problem but we can't, no one in the class has answered this. I just want to know the specific answer. Can you share your knowledge on us by answering the problem above? Thank You for sharing...=) 07-25-2007 #2 mathjoe Guest   Posts: n/a Assume we are solving for x.x^2+a(a-2x)=a^2+x(x-2a)x^2 + a^2 - 2ax = a^2 + x^2 - 2ax0 = 0The original equation is an identity. An identity is an equation that is true for all real numbers x.Answer: x is any real number. 07-25-2007 #3 ASD Guest   Posts: n/a x^2+a(a-2x)=a^2+x(x-2a)=> x^2+a^2-2ax=a^2+x^2-2ax=> x^2-x^2-2ax+2ax=0=> x(x-x+2a-2a)=0Therefore, this equation will be zero if x=0. THerefore, x=0 is a solution for this equation. 07-25-2007 #4 VeNiE Guest   Posts: n/a x^2+a(a-2x)=a^2+x(x-2a)x^2 + a^2 - 2ax = a^2 + x^2 - 2axx^2 - x^2 + a^2 - a^2 -2ax + 2ax = 0x = 0, a = 0Is this the correct answer?? 07-25-2007 #5 clairebear1951 Guest   Posts: n/a Expand both sides and getx^2 + a^2 - 2ax = a^2 + x^2 - 2axNote both sides are identical.That means it is an IDENTITY which means it is true for all values. 07-25-2007 #6 R.B.R Guest   Posts: n/a If you expand both sides, you will get the same thing x^2 + a^2 - 2ax = a^2 + x^2 - 2ax.They are both the same thing. Its like saying 1 = 1.So, it won't make any difference what value of x or a you substitute.

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