# Discriminant Sentence Examples

- This implies the vanishing of the
**discriminant**of the original form. - In this case (f, ï¿½) 1 is a perfect square, since its
**discriminant**vanishes. .-The**Discriminants****discriminant**of a homogeneous polynomial in k variables is the resultant of the k polynomials formed by differentiations in regard to each of the variables.- A binary form which has a square factor has its
**discriminant**equal to zero. - 1 X
**discriminant**of f = ao X disct. - The
**discriminant**of the product of two forms is equal to the product of theirmultiplied by the square of their resultant.**discriminants** - This follows at once from the fact that the
**discriminant**is Mara s) 2 II(/3, -fis)2{II(ar-/3$)}2. - To express the function aoa2 - _ which is the
**discriminant**of the binary quadratic aoxi -+-2a1x2x2-+a2x2 = ai =1, 1, in a symbolic form we have 2(aoa 2 -ai) =aoa2 +aGa2 -2 a1 ï¿½ al = a;b4 -}-alb? - The expression (ab) 4 properly appertains to a quartic; for a quadratic it may also be written (ab) 2 (cd) 2, and would denote the square of the
**discriminant**to a factor pres. - For the quadratic it is the
**discriminant**(ab) 2 and for ax2 the cubic the quadratic covariant (ab) 2 axbx. - The Binary Quadratic.-The complete system consists of the form itself, ax, and the
**discriminant**, which is the second transvectant of the form upon itself, viz.: (f, f') 2 = (ab) 2; or, in real coefficients, 2(a 0 a 2 a 2 1). - Calling the discriminate D, the solution of the quadratic as =o is given by the formula a: = o (a0+a12_x2 (a0x+aix2 If the form a 2 be written as the product of its linear factors p.a., the
**discriminant**takes the form -2(pq) 2. - V., established the important result that in the case of a form in n variables, the concomitants of the form, or of a system of such forms, involve in the aggregate n-1 classes of aa =5135 4 +4B8 3 p) =0, =5(135 4 - 4A 2 p 4) =0, P yield by elimination of S and p the
**discriminant**D =64B-A2. - And, dividing by y1y2...ym, the
**discriminant**of f is seen to be equal to the product of the squares of all the differences of any two roots of the equation. - The
**discriminant**of f is equal to the**discriminant**of 0, and is therefore (0, 0') 2 = R; if it vanishes both f and 0 have two roots equal, 0 is a rational factor of f and Q is a perfect cube; the cube root being equal, to a numerical factor pres, to the square root of A. - This method of solution fails when the
**discriminant**R vanishes, for then the Hessian has equal roots, as also the cubic f. - The
**discriminant**, whose vanishing is the condition that f may possess two equal roots, has the expression j 2 - 6 i 3; it is nine times the**discriminant**of the cubic resolvent k 3 - 2 ik- 3j, and has also the expression 4(1, t') 6 . - Of f=0, :and notices that they become identical on substituting 0 for k, and -f for X; hence, if k1, k2, k 3 be the roots of the resolvent -21 2 = (o + k if) (A + k 2f)(o + k 3f); and now, if all the roots of f be different, so also are those of the resolvent, since the latter, and f, have practically the same
**discriminant**; consequently each of the three factors, of -21 2, must be perfect squares and taking the square root 1 t = -' (1)ï¿½x4; and it can be shown that 0, x, 1P are the three conjugate quadratic factors of t above mentioned.