Difference between revisions of "2007 iTest Problems/Problem 23"
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== Solution == | == Solution == | ||
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Let <math>a = x^2 - 3x</math>. Substituting that in results in | Let <math>a = x^2 - 3x</math>. Substituting that in results in | ||
<cmath>a^2 + 5a + 6</cmath> | <cmath>a^2 + 5a + 6</cmath> | ||
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<cmath>(a+2)(a+3)</cmath> | <cmath>(a+2)(a+3)</cmath> | ||
<cmath>(x^2-3x+2)(x^2-3x+3)</cmath> | <cmath>(x^2-3x+2)(x^2-3x+3)</cmath> | ||
− | The first part is factorable into <math>(x-2)(x-1)</math>. The second part isn’t — and the discriminant is <math>3^2 - 4 \cdot 3 = -3</math>. With the imaginary factors found, by [[Vieta's Formulas]], the product of the non-real roots is <math> | + | The first part is factorable into <math>(x-2)(x-1)</math>. The second part isn’t — and the discriminant is <math>3^2 - 4 \cdot 3 = -3</math>. With the imaginary factors found, by [[Vieta's Formulas]], the product of the non-real roots is <math>\boxed{\textbf{(F) }3}</math>. |
==See Also== | ==See Also== |
Latest revision as of 16:48, 24 November 2018
Problem
Find the product of the non-real roots of the equation
Solution
Let . Substituting that in results in Factoring that (and substituting back) leads to The first part is factorable into . The second part isn’t — and the discriminant is . With the imaginary factors found, by Vieta's Formulas, the product of the non-real roots is .
See Also
2007 iTest (Problems) | ||
Preceded by: Problem 22 |
Followed by: Problem 24 | |
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