Anonymous
Anonymous asked in Science & MathematicsMathematics · 4 weeks ago

roots of an equation problem?

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4 Answers

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  • 4 weeks ago

    I get that a^2 + b^2 + c^2 = -2.  That's right, minus 2.

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  • ?
    Lv 7
    4 weeks ago

    Based upon the graph below,

    ..........x³ - 2x² + 3x - 4 = 0

    has only one real root at (1.651,0).

    Therefore, if we let a = 1.651,

    then b = c = 0 and

    .........a²+b²+c² = (1.651)² + 0² + 0² = 2.725801....................ANS

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    • rotchm
      Lv 7
      4 weeks agoReport

      So many errors in there, and the answer is wrong. 

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  • TomV
    Lv 7
    4 weeks ago

    If a, b, and c are roots of the equation then:

    x³ - 2x² + 3x - 4 = (x-a)(x-b)(x-c)

    = (x-a)[x²-(b+c)x + bc]

    = x³ - (a+b+c)x² + (bc +ab + ac)x - abc

    From which:

    a+b+c = 2

    ab+ac+bc = 3

    abc = 4

    Squaring the first of these:

    (a+b+c)² = 2²

    a² + b² + c² + 2(ab+ac+bc) = 4

    Substituting the known value for ab+ac+bc:

    a² + b² + c² + 2(3) = 4

    Ans: a² + b² + c² = -2

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  • rotchm
    Lv 7
    4 weeks ago

    Hint: Expand (x-a)(x-b)(x-c) and collect powers of x. Compare them to your given eqs. 

    What can you conclude of (a+b+c) = ?

    Answer this ans if need be, we will proceed. 

    If you were not anon I would detail further. 

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