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# roots of an equation problem?

### 4 Answers

- MorningfoxLv 74 weeks ago
I get that a^2 + b^2 + c^2 = -2. That's right, minus 2.

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- ?Lv 74 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

- rotchmLv 74 weeks agoReport
So many errors in there, and the answer is wrong.

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- TomVLv 74 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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- rotchmLv 74 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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