Problem: A value of xxx satisfying the equation x2+b2=(a−x)2x^{2}+b^{2}=(a-x)^{2}x2+b2=(a−x)2 is:
Answer Choices:
A. b2+a22a\dfrac{\mathrm{b}^{2}+\mathrm{a}^{2}}{2 \mathrm{a}}2ab2+a2​
B. b2−a22a\dfrac{\mathrm{b}^{2}-\mathrm{a}^{2}}{2 \mathrm{a}}2ab2−a2​
C. a2−b22a\dfrac{a^{2}-b^{2}}{2 a}2aa2−b2​
D. a−b2\dfrac{a-b}{2}2a−b​
E. a2−b22\dfrac{a^{2}-b^{2}}{2}2a2−b2​ Solution:
x2+b2=a2−2ax+x2x^{2}+b^{2}=a^{2}-2 a x+x^{2} x2+b2=a2−2ax+x2
x=a2−b22ax=\dfrac{a^{2}-b^{2}}{2 a} x=2aa2−b2​