Solve Equationswhicharereducibletoquadratic 9x^4-9x^2+2=0 Tiger ...

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((9 • (x4)) - 32x2) + 2 = 0

Step 2 :

Equation at the end of step 2 :

(32x4 - 32x2) + 2 = 0

Step 3 :

Trying to factor by splitting the middle term

3.1 Factoring 9x4-9x2+2 The first term is, 9x4 its coefficient is 9 .The middle term is, -9x2 its coefficient is -9 .The last term, "the constant", is +2 Step-1 : Multiply the coefficient of the first term by the constant 9 • 2 = 18 Step-2 : Find two factors of 18 whose sum equals the coefficient of the middle term, which is -9 .

-18	+	-1	=	-19
-9	+	-2	=	-11
-6	+	-3	=	-9	That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -6 and -3 9x4 - 6x2 - 3x2 - 2Step-4 : Add up the first 2 terms, pulling out like factors : 3x2 • (3x2-2) Add up the last 2 terms, pulling out common factors : 1 • (3x2-2) Step-5 : Add up the four terms of step 4 : (3x2-1) • (3x2-2) Which is the desired factorization

Trying to factor as a Difference of Squares :

3.2 Factoring: 3x2-2 Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)Proof : (A+B) • (A-B) = A2 - AB + BA - B2 = A2 - AB + AB - B2 = A2 - B2Note : AB = BA is the commutative property of multiplication. Note : - AB + AB equals zero and is therefore eliminated from the expression.Check : 3 is not a square !! Ruling : Binomial can not be factored as the difference of two perfect squares

Trying to factor as a Difference of Squares :

3.3 Factoring: 3x2-1 Check : 3 is not a square !! Ruling : Binomial can not be factored as the difference of two perfect squares

Equation at the end of step 3 :

(3x2 - 2) • (3x2 - 1) = 0

Step 4 :

Theory - Roots of a product :

4.1 A product of several terms equals zero.When a product of two or more terms equals zero, then at least one of the terms must be zero.We shall now solve each term = 0 separatelyIn other words, we are going to solve as many equations as there are terms in the productAny solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

4.2 Solve : 3x2-2 = 0Add 2 to both sides of the equation : 3x2 = 2 Divide both sides of the equation by 3: x2 = 2/3 = 0.667 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get: x = ± √ 2/3 The equation has two real solutions These solutions are x = ±√ 0.667 = ± 0.81650

Solving a Single Variable Equation :

4.3 Solve : 3x2-1 = 0Add 1 to both sides of the equation : 3x2 = 1 Divide both sides of the equation by 3: x2 = 1/3 = 0.333 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get: x = ± √ 1/3 The equation has two real solutions These solutions are x = ±√ 0.333 = ± 0.57735

Supplement : Solving Quadratic Equation Directly

Solving 9x4-9x2+2 = 0 directly

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

5.1 Solve 9x4-9x2+2 = 0This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using w , such that w = x2 transforms the equation into : 9w2-9w+2 = 0Solving this new equation using the quadratic formula we get two real solutions : 0.6667 or 0.3333Now that we know the value(s) of w , we can calculate x since x is √ w Doing just this we discover that the solutions of 9x4-9x2+2 = 0 are either : x =√ 0.667 = 0.81650 or : x =√ 0.667 = -0.81650 or : x =√ 0.333 = 0.57735 or : x =√ 0.333 = -0.57735

Four solutions were found :

1. x = ±√ 0.333 = ± 0.57735
2. x = ±√ 0.667 = ± 0.81650