2x² - 1 = 17 in quadratic equation

TWO SOLUTIONS:

Step-by-step explanation:

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                    2*x^2-1-(17)=0

Step by step solution : STEP 1:

Equation at the end of step 1

 (2x2 -  1) -  17  = 0

STEP 2 :

Pulling out like terms

3.1     Pull out like factors :

  2x2 - 18  =   2 • (x2 - 9)

Trying to factor as a Difference of Squares:

3.2      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 - B2

Note :  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 3

Check :  x2  is the square of  x1

Factorization is :      (x + 3)  •  (x - 3)

Equation at the end of step 2:

 2 • (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 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Equations which are never true:

3.2      Solve :    2   =  0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation:

4.3      Solve  :    x+3 = 0

Subtract  3  from both sides of the equation :

                     x = -3

Solving a Single Variable Equation:

4.4      Solve  :    x-3 = 0

Add  3  to both sides of the equation :

                     x = 3

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

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step-by-step explanation:

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