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You can combine these in many ways and so the best way to develop your intuition for the best thing to do is practice problems. A combined set of operations could be, for example,

1 a x 2 + b x = c ( 1 a x 2 + b x ) - 1 = ( c ) - 1 ( invert both sides ) a x 2 + b x 1 = 1 c a x 2 + b x = 1 c ( a x 2 + b x ) 2 = ( 1 c ) 2 ( square both sides ) a x 2 + b x = 1 c 2

Solve for x : x + 2 = x

  1. Both sides of the equation should be squared to remove the square root sign.

    x + 2 = x 2
  2. x + 2 = x 2 ( subtract x 2 from both sides ) x + 2 - x 2 = 0 ( divide both sides by - 1 ) - x - 2 + x 2 = 0 x 2 - x + 2 = 0
  3. x 2 - x + 2

    The factors of x 2 - x + 2 are ( x - 2 ) ( x + 1 ) .

  4. ( x - 2 ) ( x + 1 ) = 0
  5. We have

    x + 1 = 0

    or

    x - 2 = 0

    Therefore, x = - 1 or x = 2 .

  6. Substitute x = - 1 into the original equation x + 2 = x :

    LHS = ( - 1 ) + 2 = 1 = 1 but RHS = ( - 1 )

    Therefore LHS ≠ RHS. The sides of an equation must always balance, a potential solution that does not balance the equation is not valid. In this case the equation does not balance.

    Therefore x ≠ - 1 .

    Now substitute x = 2 into original equation x + 2 = x :

    LHS = 2 + 2 = 4 = 2 and RHS = 2

    Therefore LHS = RHS

    Therefore x = 2 is the only valid solution

  7. x + 2 = x for x = 2 only.

Solve the equation: x 2 + 3 x - 4 = 0 .

  1. The equation is in the required form, with a = 1 .

  2. You need the factors of 1 and 4 so that the middle term is + 3 So the factors are:

    ( x - 1 ) ( x + 4 )

  3. x 2 + 3 x - 4 = ( x - 1 ) ( x + 4 ) = 0

    Therefore x = 1 or x = - 4 .

  4. 1 2 + 3 ( 1 ) - 4 = 0
    ( - 4 ) 2 + 3 ( - 4 ) - 4 = 0

    Both solutions are valid.

  5. Therefore the solutions are x = 1 or x = - 4 .

Find the roots of the quadratic equation 0 = - 2 x 2 + 4 x - 2 .

  1. There is a common factor: -2. Therefore, divide both sides of the equation by -2.

    - 2 x 2 + 4 x - 2 = 0 x 2 - 2 x + 1 = 0
  2. The middle term is negative. Therefore, the factors are ( x - 1 ) ( x - 1 )

    If we multiply out ( x - 1 ) ( x - 1 ) , we get x 2 - 2 x + 1 .

  3. x 2 - 2 x + 1 = ( x - 1 ) ( x - 1 ) = 0

    In this case, the quadratic is a perfect square, so there is only one solution for x : x = 1 .

  4. - 2 ( 1 ) 2 + 4 ( 1 ) - 2 = 0 .

  5. The root of 0 = - 2 x 2 + 4 x - 2 is x = 1 .

Solving quadratic equations

  1. Solve for x : ( 3 x + 2 ) ( 3 x - 4 ) = 0
  2. Solve for x : ( 5 x - 9 ) ( x + 6 ) = 0
  3. Solve for x : ( 2 x + 3 ) ( 2 x - 3 ) = 0
  4. Solve for x : ( 2 x + 1 ) ( 2 x - 9 ) = 0
  5. Solve for x : ( 2 x - 3 ) ( 2 x - 3 ) = 0
  6. Solve for x : 20 x + 25 x 2 = 0
  7. Solve for x : 4 x 2 - 17 x - 77 = 0
  8. Solve for x : 2 x 2 - 5 x - 12 = 0
  9. Solve for x : - 75 x 2 + 290 x - 240 = 0
  10. Solve for x : 2 x = 1 3 x 2 - 3 x + 14 2 3
  11. Solve for x : x 2 - 4 x = - 4
  12. Solve for x : - x 2 + 4 x - 6 = 4 x 2 - 5 x + 3
  13. Solve for x : x 2 = 3 x
  14. Solve for x : 3 x 2 + 10 x - 25 = 0
  15. Solve for x : x 2 - x + 3
  16. Solve for x : x 2 - 4 x + 4 = 0
  17. Solve for x : x 2 - 6 x = 7
  18. Solve for x : 14 x 2 + 5 x = 6
  19. Solve for x : 2 x 2 - 2 x = 12
  20. Solve for x : 3 x 2 + 2 y - 6 = x 2 - x + 2

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Source:  OpenStax, Maths grade 10 rought draft. OpenStax CNX. Sep 29, 2011 Download for free at http://cnx.org/content/col11363/1.1
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