roots of an equation problem?
- MorningfoxLv 74 weeks ago
I get that a^2 + b^2 + c^2 = -2. That's right, minus 2.
- ?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
- 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
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
- 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.