document.write( "Question 1162485: The value of a stock, A(x), over a year long period decreased and then increased according to the
\n" ); document.write( "quadratic function 𝐴(𝑥) = 0.75𝑥
\n" ); document.write( "2 − 6𝑥 + 20, where x represents the number of months passed since
\n" ); document.write( "you invested. The value of another stock, B(x), increased linearly according to the equation
\n" ); document.write( "𝐵(𝑥) = 2.75𝑥 + 1.50 over the same year. After how long are both stocks worth the same amount?
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Algebra.Com's Answer #786290 by Boreal(15235)\"\" \"About 
You can put this solution on YOUR website!
Set the two equal
\n" ); document.write( "0.75x^2-6x+20=2.75x+1.50
\n" ); document.write( "0.75x^2-8.75x+18.5=0
\n" ); document.write( "x=(1/1.5)(8.75 +/- sqrt (76.56-55.5)); sqrt term is 4.59
\n" ); document.write( "the larger root is later in time and is (2/3)(20.22)=8.89 months\r
\n" ); document.write( "\n" ); document.write( "(the earlier intersection was at 2.77 months)\r
\n" ); document.write( "\n" ); document.write( "can also multiply through by 4 to get 3x^2-35x+74\r
\n" ); document.write( "\n" ); document.write( "\"graph%28300%2C300%2C-2%2C10%2C-10%2C30%2C0.75x%5E2-6x%2B20%2C2.75x%2B1.50%29\"
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