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11X1 T01 09 completing the square (2011)
 

11X1 T01 09 completing the square (2011)

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    11X1 T01 09 completing the square (2011) 11X1 T01 09 completing the square (2011) Presentation Transcript

    • Completing the Square
    • Completing the Squaree.g. (i ) x 2  6 x  7  0
    • Completing the Squaree.g. (i ) x 2  6 x  7  0 x2  6x  7 move the constant
    • Completing the Squaree.g. (i ) x 2  6 x  7  0 x2  6x  7 move the constant x 2  6 x  32  7  32 add half the coefficient of ‘x’ squared
    • Completing the Squaree.g. (i ) x 2  6 x  7  0 x2  6x  7 move the constant x 2  6 x  32  7  32 add half the coefficient of ‘x’ squared x 2  6 x  9  16  x  3  16 2 factorise to a perfect square
    • Completing the Squaree.g. (i ) x 2  6 x  7  0 x2  6x  7 move the constant x 2  6 x  32  7  32 add half the coefficient of ‘x’ squared x 2  6 x  9  16  x  3  16 2 factorise to a perfect square x  3  4
    • Completing the Squaree.g. (i ) x 2  6 x  7  0 x2  6x  7 move the constant x 2  6 x  32  7  32 add half the coefficient of ‘x’ squared x 2  6 x  9  16  x  3  16 2 factorise to a perfect square x  3  4 x  3  4 x  7 or x  1
    • (ii ) ax 2  bx  c  0
    • (ii ) ax 2  bx  c  0 b c x2  x   0 a a
    • (ii ) ax 2  bx  c  0 b c x2  x   0 a a b c x  x 2 a a
    • (ii ) ax 2  bx  c  0 b c x2  x   0 a a b c x  x 2 a a 2 2x2  x         b b c b     a  2a  a  2a 
    • (ii ) ax 2  bx  c  0 b c x2  x   0 a a b c x  x 2 a a 2 2x2  x         b b c b     a  2a  a  2a  2 x b  c  b 2    2a  a 4a 2 b 2  4ac  4a 2
    • (ii ) ax 2  bx  c  0 b c x2  x   0 a a b c x  x 2 a a 2 2x2  x         b b c b     a  2a  a  2a  2 x b  c  b 2    2a  a 4a 2 b 2  4ac  4a 2 b b 2  4ac x  2a 2a
    • (ii ) ax 2  bx  c  0 b c x2  x   0 a a b c x  x 2 a a 2 2x2  x         b b c b     a  2a  a  2a  2 x b  c  b 2    2a  a 4a 2 b 2  4ac  4a 2 b b 2  4ac x  2a 2a b  b 2  4ac x 2a
    • (iii ) x 2  6 x  6  0
    • (iii ) x 2  6 x  6  0  x  3 0 2
    • (iii ) x 2  6 x  6  0  x  3 3  0 2
    • (iii ) x 2  6 x  6  0  x  3 3  0 2 x  3  3  x  3  3   0
    • (iii ) x 2  6 x  6  0  x  3 3  0 2 x  3  3  x  3  3   0x  3  3 or x  3  3
    • (iii ) x 2  6 x  6  0  x  3 3  0 2 x  3  3  x  3  3   0x  3  3 or x  3  3 Exercise 1I; 1adh, 2ch, 3adg, 4bdfh, 5bdf, 6adg, 7bc, 8*