x-42-y-32-4-and-concludes-r2-4-identify-the-algebraic-error-in-the-students-work

Question

A student solves the equation: x2+y2−8x+6y+9=$$0x^2$+y^2-8x+6y+9=0 by $$writing: (x−4)2+(y+3)2=−4\left(x-4\$$right)^2$$+\left(y+3\$$right)^2=-4$$, and concludes that: r2=−$$4r^2=-4. $$Identify what the algebraic error is in the student's work.

Accepted Answer

The student made an arithmetic error on the right side when completing the square. Adding the constants gives −9+16+9=16-9+16+9=16, so the correct form is (x−4)2+(y+3)2=16\left(x-4\$$right)^2$$+\left(y+3\$$right)^2=16.$$

Answer

The student should have divided the entire equation by 22 before completing the square.

Answer

The student completed the square incorrectly; the binomials (x−4)2\left(x-4\$$right)^2$$ and (y+3)2\left(y+3\$$right)^2$$ are not valid.

Answer

There is no algebraic error; circles with a negative $$r2r^2$$ are valid within the real plane.

A student solves the equation: x2+y2−8x+6y+9=0x^2+y^2-8x+6y+9=0 by writing: (x−4)2+(y+3)2=−4\(\left\)(x-4\(\right\))^2+\(\left\)(y+3\(\right\))^2=-4, and concludes that: r2=−4r^2=-4. Identify what the algebraic error is in the student's work.

A student solves the equation: $x^{2} + y^{2} - 8 x + 6 y + 9 = 0$ by writing: ${(x - 4)}^{2} + {(y + 3)}^{2} = - 4$ , and concludes that: $r^{2} = - 4$ . Identify what the algebraic error is in the student's work.