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Solving Algebraic Equations

Solving Algebraic Equations.

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Solving Algebraic Equations

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  1. Solving Algebraic Equations There are many types of equations starting with linear and including rational, Quadratic with and without factoring, absolute value, cube root and square root. In all these types there are basic rules; get the variables to one side of the equation, remove all the numbers and operations from the variable side, and check for extraneous solutions.

  2. Lets start with linear equations. First of all what does it mean by a linear equation? The variable(s) have a degree of 1 For example: 3x + 5 = 7 variable is on one side and combined already 3x = 2 start removing numbers and operations x = 2/3 until the variable is by itself Example Practice: 3(2 – x) = 2x – 1 8x – (2x + 1) = 3x – 10 6 – 3x = 2x – 1 6 = 5x – 1 7 = 5x 7/5 = x

  3. Solving Rational Equations This means we are finding the value of x that will make the equation true. We are not just simplifying to a simpler expression.

  4. Practice: 2 rationals equal to each other cross multiply. 2 rationals equal to each other cross multiply.

  5. 3x 3x 3x 2x(x+1) 2x(x+1) 2x(x+1) More then 2 rational equations More then 2 rational equations LCD: 3x LCD: 2x(x+1)

  6. Practice:

  7. * An absolute value can never = a negative value!! However, the value inside the absolute value can be negative it just becomes positive when you apply the absolute value to it.

  8. EXAMPLE 2 Solve an absolute value equation Solve|2x – 3 | = 7. SOLUTION | 2x – 3 | = 7 Write original equation. 2x – 3 = 7or –(2x –3) = 7 2x – 3 = -7 When it is taken out of the absolute value the expression can equal 7 or –7 . +3 +3 +3 +3 2x = 10or 2x = -4 Add 3 to each side. 2 2 2 2 x = 5or x = - 2 Divide each side by 2.

  9. GUIDED PRACTICE Solve the equation. Check for extraneous solutions. The absolute value must be alone on one side before you can split it up into equivalent equations. 2|x – 3| + 4 = 10 SOLUTION 2| x – 3| + 4 = 10 Write original equation. 2| x – 3| = 6 | x – 3| = 3 Simplify: get the absolute value by itself. -(x – 3) = 3 x – 3= – 3 or x – 3 = 3 Write equivalent equations. x = -3 + 3orx = 3 + 3 Solve forx. x = 0 or x = 6

  10. Solve the equation. Check for extraneous solutions. An extraneous solution is a solution that appears to be correct but is not. Must check to find out. |9 – 2x| = 10 +3x SOLUTION |9 – 2x| = 10 +3x Write original equation. -(9 – 2x) = 10 + 3x 9 – 2x = 10 + 3xor 9 – 2x = -10 – 3x Write equivalent equations. -1 = 5xorx = -19 Solve forx. x = – 1/5 orx = -19 Simplify.

  11. 9 2/5= 9 2/5 47= - 47 ANSWER The solution is -1/5. Reject -19 because it is an extraneous solution. EXAMPLE 3 Check for extraneous solutions Check the apparent solutions to see if either is extraneous. CHECK |9 – 2x| = 10 +3x |9 – 2x| = 10 +3x ? ? |9 – 2(-1/5)| = 10 +3(-1/5) |9 – 2(-19)| = 10 +3(-19)

  12. Practice: 1.) |-2x| = 8 2.) |3x – 1| = 2 3.) |x2 + 3x - 2| = 2

  13. Solving by factoring 1.) Make sure the equation is equal to 0. 2.) Factor 3.) Set each factor = to zero and solve each factor seperately. x2 – 8x + 12 = 0 x2 – 6x = 0 -3x + 28 = x2 (x – 6)(x – 2) x(x – 6) x2 + 3x – 28 = 0 x = 0 x – 6 = 0 x = 6 x – 6 = 0 + 6 + 6 x = 6 x – 2 = 0 + 2 +2 x = 2 (x +7)(x – 4) x = -7 x = 4

  14. Practice: 4z3 – 8z2 = 0 4x2 + 9 = 12x

  15. Solving with Square Roots Get the squared part by itself on one side of the =. Take the square root of both sides leaving one side without the square. The answer side will get a plus and a minus sign. If there is anything else to get rid of to get the variable by itself finish with that.

  16. Examples

  17. Completing the Square Do you remember when we factored sometimes the factors came out to be the same? For example x2 + 6x + 9 (x + 3) (x + 3) Well we can write that as (x +3)2 and then we can solve using square roots.

  18. This only works if the last number is just right so if it is not the right number we have to make it the right number. +25 10/2 = 52 + 9 6/2 = 32 x2 + 10x _____ x2 + 6x _____ (x + 5)2 +36 12/2 = 62 + 169/4 (13/2)2 x2 – 12x _____ x2 – 13x _____ (x – 6)2 What is the pattern to find the value that makes the trinomial a perfect square? (b/2)2

  19. Now lets use this to solve problems: If you add to one side you have to add to the other + 1 x2 + 2x _____ = 10 + 1 x2 + 8x -9 = 0 (x + 1)2 = 11 x2 + 8x = 9 +16 +16 (x+4)2 = 25

  20. Practice: What number should be added to complete the square? x2 – 2/3 x Solve the equation by completing the square. x2 – 6x = 13

  21. Quadratic Formula You can use this formula to find the solutions(roots) to a quadratic equation. This formula can be broken up into 2 parts: b2 – 4ac is called the Discriminant. It can tell you what the roots look like.

  22. Find the Discriminant and describe the type of roots: y = 3x2 + 5x – 4 a = 3 b = 5 c = -4 (5)2 – 4(3)(-4) = 25 + 48 = 73 2 real roots Solve for the roots. Practice finding the discriminant 5x2 + 20x + 21 = 0

  23. Solving the roots: y = 2x2 – 3x + 6 a = 2 b = -3 c = 6 Discriminant 9 – 48 = - 39 Practice solving: x2 + 4x + 2 = 0 4t2 + t + 1 = 0

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