Consider a quadratic equation of the form x + bx +c = 0, where b and c are rational numbers. Fill in the blanks in the following proof that if one solution is rational, then the other solution is also rational. Proof: Suppose x + bx +c = 0 is any quadratic equation where b and c are rational numbers, and suppose one solution r is rational. Call the other solution s. Then x + bx +c = (x -r)(x - s). Multiply out (x - r)(x -s) and set it equal to x + bx + c to obtain x2 + bx +c H x+ Equate coefficients and solve for s in terms of b and r to obtain s = -6 -r Since b and r vv are rational vv and since differences of rational numbers v are rational, we conclude that -b -r is rational, and so s is rational.
Consider a quadratic equation of the form x + bx +c = 0, where b and c are rational numbers. Fill in the blanks in the following proof that if one solution is rational, then the other solution is also rational. Proof: Suppose x + bx +c = 0 is any quadratic equation where b and c are rational numbers, and suppose one solution r is rational. Call the other solution s. Then x + bx +c = (x -r)(x - s). Multiply out (x - r)(x -s) and set it equal to x + bx + c to obtain x2 + bx +c H x+ Equate coefficients and solve for s in terms of b and r to obtain s = -6 -r Since b and r vv are rational vv and since differences of rational numbers v are rational, we conclude that -b -r is rational, and so s is rational.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Consider a quadratic equation of the form x2 + bx + c = 0, where b and c are rational numbers. Fill in the blanks in the following proof that if one solution is rational, then the other solution is also rational.
Proof: Suppose x? + bx + c = 0 is any quadratic equation where b and c are rational numbers, and suppose one solution r is rational. Call the other solution s. Then x2 + bx + c = (x - r)(x - s). Multiply out (x - r)(x - s) and set it equal to x? + bx + c to obtain
x2 + bx + c = x2 +
. Equate coefficients and solve for s in terms of b and r to obtain s =
-6 -r
. Since b and r
are rational
and since differences of rational numbers v
are rational,
rs
we conclude that -b -r
is rational, and so s is rational.
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