Solve Equationswhicharereducibletoquadratic X^4-5x^2-36=0 Tiger ...

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((x4) - 5x2) - 36 = 0

Step 2 :

Trying to factor by splitting the middle term

2.1 Factoring x4-5x2-36 The first term is, x4 its coefficient is 1 .The middle term is, -5x2 its coefficient is -5 .The last term, "the constant", is -36 Step-1 : Multiply the coefficient of the first term by the constant 1 • -36 = -36 Step-2 : Find two factors of -36 whose sum equals the coefficient of the middle term, which is -5 .

-36	+	1	=	-35
-18	+	2	=	-16
-12	+	3	=	-9
-9	+	4	=	-5	That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -9 and 4 x4 - 9x2 + 4x2 - 36Step-4 : Add up the first 2 terms, pulling out like factors : x2 • (x2-9) Add up the last 2 terms, pulling out common factors : 4 • (x2-9) Step-5 : Add up the four terms of step 4 : (x2+4) • (x2-9) Which is the desired factorization

Polynomial Roots Calculator :

2.2 Find roots (zeroes) of : F(x) = x2+4Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0 Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integersThe Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading CoefficientIn this case, the Leading Coefficient is 1 and the Trailing Constant is 4. The factor(s) are: of the Leading Coefficient : 1 of the Trailing Constant : 1 ,2 ,4 Let us test ....

P	Q	P/Q	F(P/Q)	Divisor
-1	1	-1.00	5.00
-2	1	-2.00	8.00
-4	1	-4.00	20.00
1	1	1.00	5.00
2	1	2.00	8.00
4	1	4.00	20.00

Polynomial Roots Calculator found no rational roots

Trying to factor as a Difference of Squares :

2.3 Factoring: x2-9 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 : 9 is the square of 3Check : x2 is the square of x1 Factorization is : (x + 3) • (x - 3)

Equation at the end of step 2 :

(x2 + 4) • (x + 3) • (x - 3) = 0

Step 3 :

Theory - Roots of a product :

3.1 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 :

3.2 Solve : x2+4 = 0Subtract 4 from both sides of the equation : x2 = -4 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get: x = ± √ -4 In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1 Accordingly, √ -4 = √ -1• 4 = √ -1 •√ 4 = i • √ 4 Can √ 4 be simplified ?Yes! The prime factorization of 4 is 2•2 To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).√ 4 = √ 2•2 = ± 2 • √ 1 = ± 2 The equation has no real solutions. It has 2 imaginary, or complex solutions. x= 0.0000 + 2.0000 i x= 0.0000 - 2.0000 i

Solving a Single Variable Equation :

3.3 Solve : x+3 = 0Subtract 3 from both sides of the equation : x = -3

Solving a Single Variable Equation :

3.4 Solve : x-3 = 0Add 3 to both sides of the equation : x = 3

Supplement : Solving Quadratic Equation Directly

Solving x4-5x2-36 = 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 :

4.1 Solve x4-5x2-36 = 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 : w2-5w-36 = 0Solving this new equation using the quadratic formula we get two real solutions : 9.0000 or -4.0000Now 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 x4-5x2-36 = 0 are either : x =√ 9.000 = 3.00000 or : x =√ 9.000 = -3.00000 or : x =√-4.000 = 0.0 + 2.00000 i or : x =√-4.000 = 0.0 - 2.00000 i

Four solutions were found :

1. x = 3
2. x = -3
3. x= 0.0000 - 2.0000 i
4. x= 0.0000 + 2.0000 i