document.write( "Question 160427: How to show that the equation x^2 + px -q^2 = 0 are real for any real numbers p & q? Thanks \n" ); document.write( "

Algebra.Com's Answer #118335 by jim_thompson5910(35256) $\"\"$ $\"About$

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From $\"x%5E2+%2B+px+-q%5E2+=+0\"$ we can see that $\"a=1\"$ , $\"b=p\"$ , and $\"c=-q%5E2\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"D=b%5E2-4ac\"$ Start with the discriminant formula.\r
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\n" ); document.write( "\n" ); document.write( " $\"D=p%5E2-4%281%29%28-q%5E2%29\"$ Plug in $\"a=1\"$ , $\"b=p\"$ , and $\"c=-q%5E2\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"D=p%5E2%2B4q%5E2\"$ Multiply \r
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\n" ); document.write( "\n" ); document.write( "So for any value of \"p\", the value of $\"p%5E2\"$ is nonnegative. So $\"p%5E2\"$ is always nonnegative (ie it is zero or a positive number).\r
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\n" ); document.write( "\n" ); document.write( "Also for any value of \"q\", the value of $\"q%5E2\"$ is nonnegative. So by extension, $\"4q%5E2\"$ is always nonnegative.\r
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\n" ); document.write( "\n" ); document.write( "So this means that $\"p%5E2%2B4q%5E2\"$ is always nonnegative which means that $\"p%5E2%2B4q%5E2%3E=0\"$ and $\"D%3E=0\"$ . Since the discriminant D is greater than or equal to zero, this means that the equation $\"x%5E2+%2B+px+-q%5E2+=+0\"$ will have real solutions for any values of \"p\" and \"q\" \n" ); document.write( "

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