document.write( "Question 88512: use the quadratic formula to solve z^2-8z=-5 \n" ); document.write( "
Algebra.Com's Answer #64350 by jim_thompson5910(35256)\"\" \"About 
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\"z%5E2-8z=-5\"\r
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\n" ); document.write( "\n" ); document.write( "\"z%5E2-8z%2B5=0\" Add 5 to both sides\r
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\n" ); document.write( "\n" ); document.write( "Now let's use the quadratic formula to solve for z:\r
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\n" ); document.write( "\n" ); document.write( "Starting with the general quadratic\r
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\n" ); document.write( "\n" ); document.write( "\"az%5E2%2Bbz%2Bc=0\"\r
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\n" ); document.write( "\n" ); document.write( "the general solution using the quadratic equation is:\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29\"\r
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\n" ); document.write( "\n" ); document.write( "So lets solve \"z%5E2-8%2Az%2B5=0\" ( notice \"a=1\", \"b=-8\", and \"c=5\")\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%28--8+%2B-+sqrt%28+%28-8%29%5E2-4%2A1%2A5+%29%29%2F%282%2A1%29\" Plug in a=1, b=-8, and c=5\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B-+sqrt%28+%28-8%29%5E2-4%2A1%2A5+%29%29%2F%282%2A1%29\" Negate -8 to get 8\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B-+sqrt%28+64-4%2A1%2A5+%29%29%2F%282%2A1%29\" Square -8 to get 64\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B-+sqrt%28+64%2B-20+%29%29%2F%282%2A1%29\" Multiply \"-4%2A5%2A1\" to get \"-20\"\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B-+sqrt%28+44+%29%29%2F%282%2A1%29\" Combine like terms in the radicand (everything under the square root)\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B-+2%2Asqrt%2811%29%29%2F%282%2A1%29\" Simplify the square root\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B-+2%2Asqrt%2811%29%29%2F2\" Multiply 2 and 1 to get 2\r
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\n" ); document.write( "\n" ); document.write( "So now the expression breaks down into two parts\r
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\n" ); document.write( "\n" ); document.write( "\"z+=+%288+%2B+2%2Asqrt%2811%29%29%2F2\" or \"z+=+%288+-+2%2Asqrt%2811%29%29%2F2\"\r
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\n" ); document.write( "\n" ); document.write( "Which approximate to\r
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\n" ); document.write( "\n" ); document.write( "\"z=7.3166247903554\" or \"z=0.6833752096446\"\r
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\n" ); document.write( "\n" ); document.write( "So our solutions are:\r
\n" ); document.write( "\n" ); document.write( "\"z=7.3166247903554\" or \"z=0.6833752096446\"\r
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\n" ); document.write( "\n" ); document.write( "Notice when we graph \"x%5E2-8%2Ax%2B5\" (just replace z with x), we get:\r
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\n" ); document.write( "\n" ); document.write( "when we use the root finder feature on a calculator, we find that \"x=7.3166247903554\" and \"x=0.6833752096446\".So this verifies our answer
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