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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 11

Solve each equation in Exercises 1 - 14 by factoring. 2x(x−3)=5x2−7x2x(x - 3) = 5x^2 - 7x

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1
Start by expanding the left side of the equation: multiply 2x by each term inside the parentheses to get \(2x \cdot x\) and \(2x \cdot (-3)\), which gives \(2x^{2} - 6x\).
Rewrite the equation with the expanded left side: \(2x^{2} - 6x = 5x^{2} - 7x\).
Bring all terms to one side to set the equation equal to zero. Subtract \$5x^{2}$ and add \$7x$ to both sides: \(2x^{2} - 6x - 5x^{2} + 7x = 0\).
Combine like terms: \(2x^{2} - 5x^{2} = -3x^{2}\) and \(-6x + 7x = x\), so the equation becomes \(-3x^{2} + x = 0\).
Factor the resulting expression by taking out the greatest common factor (GCF), which is \(x\): \(x(-3x + 1) = 0\). Then, set each factor equal to zero to find the solutions.

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