document.write( "Question 139569: What type of solution do you get for quadratic equations where D < 0? Give reasons for your answer. \r
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Algebra.Com's Answer #101760 by solver91311(24713) $\"\"$ $\"About$

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I presume by 'D' you mean the determinant in the quadratic equation. If D is less than zero, then you are faced with the problem of taking the square root of a negative number. If you think about that for a moment, you will realize it is an impossible situation. Taking a square root means finding two equal factors for a number and if the factors are equal, they must have the same sign. If two numbers with the same sign are multiplied together, you always get a positive result. So now what?\r
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\n" ); document.write( "\n" ); document.write( "Mathematicians invented a special number, called an imaginary number, i, that is defined thusly: $\"i%5E2=-1\"$ . Turns out that this is the only imaginary number needed because any negative number can be expressed as the product of -1 and the opposite of the number. -2 is the same as (-1)*2, for example. Since $\"sqrt%28ab%29=sqrt%28a%29sqrt%28b%29\"$ , if you had to deal with $\"sqrt%28-2%29\"$ , you could change it to $\"sqrt%28-1%29sqrt%282%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "What happens when the discriminant is negative, or less than 0, is this:\r
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\n" ); document.write( "\n" ); document.write( " $\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"$ where $\"D=b%5E2-4ac%3C0\"$ becomes\r
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\n" ); document.write( "\n" ); document.write( " $\"x+=+%28-b+%2B-+sqrt%28+%28-1%29%28-D%29+%29%29%2F%282%2Aa%29+\"$ but we say that $\"sqrt%28-1%29=i\"$ so:\r
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\n" ); document.write( "\n" ); document.write( " $\"x+=+%28-b+%2B-+i%2Asqrt%28%28-D%29+%29%29%2F%282%2Aa%29+\"$ (Note that if D < 0, -D > 0)\r
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\n" ); document.write( "\n" ); document.write( "So D < 0 means that you will have a conjugate pair of complex roots of the form $\"alpha%2B-beta%2Ai\"$ where $\"alpha\"$ and $\"beta\"$ are real numbers and i is defined by $\"i%5E2=-1\"$ Furthermore, solutions to the general quadratic with D < 0 will be $\"alpha%2B-beta%2Ai\"$ where $\"alpha=-b%2F2a\"$ and $\"beta=sqrt%28-D%29%2F2a\"$ \r
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\n" ); document.write( "\n" ); document.write( "Note that complex roots ALWAYS come in conjugate pairs. If you find a complex root to a polynomial equation that is $\"alpha%2Bbeta%2Ai\"$ , then it is guaranteed that $\"alpha-beta%2Ai\"$ is also a root. This is also true for higher degree polynomial equations. It is no coincidence that quadratic (degree 2) equations have 2 roots. In fact, cubics (3rd degree) equations have 3 roots, quartics (4th degree) equations have 4 roots, and nth degree equations have n roots. As a consequence of the fact that complex roots must come in pairs, equations of odd degree (3rd, 5th, etc.) MUST have at least one real root.\r
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\n" ); document.write( "\n" ); document.write( "Wait a minute, you say! Yes, you have solved quadratics where you only got one answer. In fact, you may have been told that if D=0 there is only 1 real root. To that I say, 'not so fast, mathematics breath'.\r
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\n" ); document.write( "\n" ); document.write( "If D=0 then you have a perfect square trinomial, something like $\"x%5E2-4x%2B4=0\"$ which factors to $\"%28x-2%29%28x-2%29=0\"$ . [You can prove that by applying the quadratic equation to $\"x%5E2%2B2px%2Bp%5E2\"$ where $\"a=1\"$ , $\"b=2p\"$ , and $\"c=p%5E2\"$ . D will come out to be $\"4p%5E2-4p%5E2\"$ ]\r
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\n" ); document.write( "\n" ); document.write( "Just because the factors are identical doesn't mean there aren't 2 of them. And if there are 2 factors, there are also 2 roots. They just happen to be identical. Some mathematicians use the language 'one real root with a multiplicity of 2' Same thing as far as I'm concerned. \n" ); document.write( "

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