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 x2 + 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 = x? + ( Equate coefficients and solve for s in terms of b and c to obtain s = Since ---Select- v are --Select-- V and since -Select- v are rational, we conclude that ---Select-- v is rational, and so sis rational.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
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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 x2 + 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 = x? + (
Equate coefficients and solve for s in terms of b and c to obtain s =
Since ---Select- v are --Select-- V and since -Select-
v are rational, we conclude that ---Select-- v is
rational, and so sis rational.
Transcribed Image Text: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 x2 + 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 = x? + ( Equate coefficients and solve for s in terms of b and c to obtain s = Since ---Select- v are --Select-- V and since -Select- v are rational, we conclude that ---Select-- v is rational, and so sis rational.
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