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One particular solution is x = 0, y = 0, z = 0. Two other solutions are x = 3, y = 6, z = 1, and x = 8, y = 9, z = 2. There is a unique plane in three-dimensional space which passes through the three points with these coordinates, and this plane is the set of all points whose coordinates are solutions of the equation.
The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
To begin solving, we multiply each side of the equation by the least common denominator of all the fractions contained in the equation. In this case, the least common denominator is ( x − 2 ) ( x + 2 ) {\displaystyle (x-2)(x+2)} .
Because (a + 1) 2 = a, a + 1 is the unique solution of the quadratic equation x 2 + a = 0. On the other hand, the polynomial x 2 + ax + 1 is irreducible over F 4, but it splits over F 16, where it has the two roots ab and ab + a, where b is a root of x 2 + x + a in F 16. This is a special case of Artin–Schreier theory.
For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0.
How to Solve It suggests the following steps when solving a mathematical problem: . First, you have to understand the problem. [2]After understanding, make a plan. [3]Carry out the plan.
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If an equation P(x) = 0 of degree n has a rational root α, the associated polynomial can be factored to give the form P(X) = (X – α)Q(X) (by dividing P(X) by X – α or by writing P(X) – P(α) as a linear combination of terms of the form X k – α k, and factoring out X – α. Solving P(x) = 0 thus reduces to solving the degree n – 1 ...