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Trigonometric identities may help simplify the answer. [ 1 ] [ 2 ] Like other methods of integration by substitution, when evaluating a definite integral, it may be simpler to completely deduce the antiderivative before applying the boundaries of integration.
For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement. It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1.
the roots of this irreducible polynomial can be calculated as [5] 1 ± 2 1 / 6 , 1 ± − 1 ± 3 i 2 1 / 3 . {\displaystyle 1\pm 2^{1/6},1\pm {\frac {\sqrt {-1\pm {\sqrt {3}}i}}{2^{1/3}}}.} Even in the case of quartic polynomials , where there is an explicit formula for the roots, solving using the decomposition often gives a simpler form.
For example, the statement = is true if is either 2 or −2 and false otherwise. [26] Equations with variables can be divided into identity equations and conditional equations. Identity equations are true for all values that can be assigned to the variables, such as the equation 2 x + 5 x = 7 x {\displaystyle 2x+5x=7x} .
Since taking the square root is the same as raising to the power 1 / 2 , the following is also an algebraic expression: 1 − x 2 1 + x 2 {\displaystyle {\sqrt {\frac {1-x^{2}}{1+x^{2}}}}} An algebraic equation is an equation involving polynomials , for which algebraic expressions may be solutions .
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.
Simplifying this further gives us the solution x = −3. It is easily checked that none of the zeros of x ( x + 1)( x + 2) – namely x = 0 , x = −1 , and x = −2 – is a solution of the final equation, so no spurious solutions were introduced.
In mathematics, an expansion of a product of sums expresses it as a sum of products by using the fact that multiplication distributes over addition. Expansion of a polynomial expression can be obtained by repeatedly replacing subexpressions that multiply two other subexpressions, at least one of which is an addition, by the equivalent sum of products, continuing until the expression becomes a ...