Roots of the Quadratic EquationDate: 4/10/96 at 14:23:37 From: kelli rostkowski Subject: Algebra: quadratic equations Would you explain to me why a quadratic equation cannot have one irrational or one imaginary root. Date: 4/27/96 at 19:42:54 From: Doctor Steven Subject: Re: Algebra: quadratic equations I'm assuming you mean a quadratic equation with rational coefficients. The reason why it can't have two complex roots is that if one complex number is a root, then its conjugate must be a root, in order to make the constant at the end of the quadratic a real number. The reason it can't have just one irrational root is that an irrational multiplied by a rational is still irrational, so the constant at the end of the quadratic would be irrational. To see this, say h and g are the roots and p(x) is the quadratic: p(x) = x^2 + a*x + b But p(x) is: (x - h)(x - g). So (x - h)(x - g) = x^2 + a*x + b Simplify the left side to get: x^2 + (-h - g)*x + h*g = x^2 + a*x + b. So -h - g = a and h*g = b. Both a and b must be rational. If h is irrational and g isn't, then -h - g is irrational, and h*g must also be irrational. But this can't happen, so one root can't be irrational and the other not. If h is complex, and g is not its conjugate, then -h - g is still complex, and h*g is also complex then. But this can't happen, so one root can't be complex and the other not its conjugate (which is also complex). Hope this helps. -Doctor Steven, The Math Forum |
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