There, We presented the Derivation Of Quadratic Formula in a lucid way. An Example in applying the formula is given. Another problem for practice is also given.

Solve the following equation using Quadratic Formula

2x^{2} + 3x - 3 = 0

Solution of Example 1 of Steps of Quadratic Formula :

To solve 2x^{2} + 3x - 3 = 0 using Quadratic Formula Comparing this equation with ax^{2} + bx + c = 0, we get a = 2, b = 3 and c = -3 We know by Quadratic Formula, x = {(-b) ± √(b^{2} - 4ac)}⁄2a Applying this Quadratic Formula here, we get x = {(-b) ± √(b^{2} - 4ac)}⁄2a = [ (-3) ± √{(3)^{2} - 4(2)(-3)}]⁄2(2) = [ (-3) ± √{9 + 24}]⁄4 = [-3 ± √(33)]⁄4 Ans.

Example 1 of Steps of Quadratic Formula is thus solved.

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Solve the following equation using Quadratic Formula 2x^{4} - 9x^{3} + 14x^{2} - 9x + 2 = 0

Solution of Example 2 of Steps of Quadratic Formula :

The given equation in descending powers of x is 2x^{4} - 9x^{3} + 14x^{2} - 9x + 2 = 0 This is an equation of the form ax^{4} + bx^{3} + cx^{2} + bx + a = 0 in which the coefficients of terms equidistant from first and last are equal, is called a RECIPROCAL EQUATION. By dividing this reciprocal equationwith x^{2} and with a proper substitution, the reciprocal equationcan be reduced to a Quadratic Equation. Dividing the equation by x^{2}, we get 2x^{2} - 9x + 14 - 9x^{-1} + 2x^{-2} = 0 ⇒ 2x^{2} + 2⁄x^{2} - 9x - 9⁄x + 14 = 0 ⇒ 2(x^{2} + 1⁄x^{2}) -9(x + 1⁄x) + 14 = 0 Putting x + 1⁄x = t, x^{2} + 1⁄x^{2} = (x + 1⁄x)^{2} - 2 = t^{2} - 2 ∴ the given equation becomes 2(t^{2} - 2) -9t + 14 = 0 ⇒ 2t^{2} - 4 - 9t + 14 = 0 ⇒ 2t^{2} - 9t + 10 = 0 Comparing this equation with at^{2} + bt + c = 0, we get a = 2, b = -9 and c = 10 We know by Quadratic Formula, t = {-b ± √(b^{2} - 4ac)}⁄2a Applying this Quadratic Formula here, we get t = {-b ± √(b^{2} - 4ac)}⁄2a = [-(-9) ± √{(-9)^{2} - 4(2)(10)}]⁄2(2) = [ (+9) ± √{81 - 80}]⁄4 = [ (9) ± √{1}]⁄4 = [ (9) ± 1]⁄4 = (9 + 1)⁄4, (9 - 1)⁄4 = 10⁄4, 8⁄4 = 5⁄2, 2

But t = x + 1⁄x

Taking t = 5⁄2, t = 5⁄2 ⇒ x + 1⁄x = 5⁄2 Multiplying both sides with 2x, we get 2x^{2} + 2 = 5x ⇒ 2x^{2} - 5x + 2 = 0Comparing this equation with ax^{2} + bx + c = 0, we get a = 2, b = -5 and c = 2 We know by Quadratic Formula, x = {-b ± √(b^{2} - 4ac)}⁄2a Applying this Quadratic Formula here, we get x = {-b ± √(b^{2} - 4ac)}⁄2a = [ -(-5) ± √{(-5)^{2} - 4(2)(2)}]⁄2(2) = [ (+5) ± √{25 - 16}]⁄4 = [ (5) ± √{9}]⁄4 = [ (5) ± 3]⁄4 = (5 + 3)⁄4, (5 - 3)⁄4 = 8⁄4, 2⁄4 = 2, 1⁄2

Taking t = 2, t = 2 ⇒ x + 1⁄x = 2 Multiplying both sides with x, we get x^{2} + 1 = 2x ⇒ x^{2} - 2x + 1 = 0 Comparing this equation with ax^{2} + bx + c = 0, we get a = 1, b = -2 and c = 1 We know by Quadratic Formula, x = {-b ± √(b^{2} - 4ac)}⁄2a Applying this Quadratic Formula here, we get x = {-b ± √(b^{2} - 4ac)}⁄2a = [ -(-2) ± √{(-2)^{2} - 4(1)(1)}]⁄2(1) = [ (+2) ± √{4 - 4}]⁄2 = [ (2) ± √{0}]⁄2 = [ (2) ± 0]⁄2 = (2 + 0)⁄2, (2 - 0)⁄2 = 2⁄2, 2⁄2 = 1, 1

Thus the solution of the given equation is x = { 1, 1, 2, 1⁄2 } Ans.

Example 2 of Steps of Quadratic Formula is thus solved.

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Solve the following equations using Steps of Quadratic Formula 12x^{2} + 3x - 99 = 0

Solve the following equation using Quadratic Formula 6x^{4} - 35x^{3} + 62x^{2} - 35x + 6 = 0

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