document.write( "Question 523322: suppose the roots of ax^2 + bx + c =0 are r and s. which one of the following has roots ar + b and as + b ? \r
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document.write( "(A) x^2 - bx - ac =0\r
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document.write( "(B) x^2 - bx + ac =0\r
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document.write( "(C) x^2 + 3bx + ca + 2b^2 =0\r
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document.write( "(D) x^2 + 3bx - ca + 2b^2 =0\r
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document.write( "(E) not given \r
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Algebra.Com's Answer #347388 by richard1234(7193)![]() ![]() You can put this solution on YOUR website! Recall that, if z is a root of a polynomial, then x-z is a factor of the polynomial.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Hence, we can write our new polynomial equation as (x - (ar + b))(x - (as + b)) = 0. Expanding the LHS, this yields\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Recall that by Vieta's formulas, r+s = -b/a and rs = c/a. We can substitute these expressions in:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "b^2 cancels, so you will see that this equation is the same as in answer choice B. \n" ); document.write( " |