Solving Cubic Polynomials - SHSU

Solving Cubic The general solution to the quadratic equationThere are four steps to finding the zeroes of a quadratic First divide by the leading term, making the Then, givenx2+a1x+a0, substitutex=y a12to obtain an equation without the linear term.(This is the depressed equation.)3. Solve then foryas a square root . (Remember to use both signs of the square root .)4. Once this is done, recoverxusing the fact thatx=y example, let s solve2x2+ 7x 15 = , we divide both sides by 2 to create an equation with leading term equal to one:x2+72x 152= replacexbyx=y a12=y 74to obtain:y2=16916 Solve fory:y=134or 134 Then, Solving back forx, we havex=32or method is equivalent to completing the square and is the steps taken in developing the much-memorized quadratic fo

3.Solve then for yas a square root. (Remember to use both signs of the square root.) 4.Once this is done, recover xusing the fact that x= y a 1 2. For example, let’s solve 2x2 + 7x 15 = 0: First, we divide both sides by 2 to create an equation with leading term equal to one: x2 + 7 2 x 15 2 = 0: Then replace xby x= y a 1 2 = y 7 4 to obtain ...

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  Square, Root, Square roots, Shsu

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