Math, asked by pardhumundru1423, 10 months ago

Roots of x3 + ax2 + bx + c = 0 are
cosines of angles of an acute triangle, then
the value of a 2b – 2c is
2

Answers

Answered by RvChaudharY50

Question :- Roots of x³ + ax² + bx + c = 0 are cosines of angles of an acute triangle, then the value of a² - 2b - 2c is ?

Solution :-

We know that, the cubic polynomial P(x) =ax³ + bx² + cx + d = 0 ,a ≠ 0. If α, β and γ are zeros of P(x),

Then :-

α + β + γ = (-b)/a
αβ + βγ + αγ = c/a
αβγ = (-d)/a

Comparing the given polynomial x³ + ax² + bx + c = 0 with ax³ + bx² + cx + d = 0 , we get,

a = 1
b = a
c = b
d = c

Than,

Also, roots of given Polynomial are cosA , cosB and cosC .

Therefore,

cosA + cosB + cosC = (-b)/a = (-a)/1 = (-a) .
cosA*cosB + cosB*cosC + cosA*cosC = c/a = b/1 = b.
cosA * cosB * cosC = (-d)/a = (-c)/1 = (-c) .

Putting all these value now, we get,

→ a² - 2b - 2c

→ (cosA + cosB + cosC)² - 2(cosA*cosB + cosB*cosC + cosA*cosC) - 2{(-1) * cosA * cosB * cosC }

using (a + b + c)² = a² + b² + c² + 2(ab + bc + ac) , we get,

→ {cos²A + cos²B + cos²C + 2(cosA*cosB + cosB*cosC + cosA*cosC)} - 2(cosA*cosB + cosB*cosC + cosA*cosC) + 2(cosA * cosB * cosC)

→ cos²A + cos²B + cos²C + 2(cosA * cosB * cosC)

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Now, Lets Try to Solve second Part :-

→ 2(cosA * cosB * cosC)

→ (2 * cosA * cosB) * cosC

using 2*cosC*cosD = cos(C + D) + cos(C - D) , we get,

→ {cos(A + B) + cos(A - B)} * cosC

By angle sum Property of ∆,

A + B + C = π
(A + B) = (π - C)

Putting, we get,

→ { cos(π - C) + cos(A - B) } * cosC

→ cos(π - C)*cosC + cos(A - B)*cosC

using cos(π - A) = - (cosA) now,

→ -(cosC) * cosC + cos(A - B)*cos{π - (A + B)}

→ - cos²C + cos(A - B) * (-1)cos(A + B)

→ - cos²C - cos(A - B) * cos(A + B)

using cos(A - B) * cos(A + B) = cos²A - sin²B , we get,

→ - cos²C - [cos²A - sin²B ]

→ - cos²C - cos²A + sin²B

using sin²B = (1 - cos²B) finally, we get,

→ - cos²C - cos²A + (1 - cos²B)

→ [ 1 - cos²C - cos²A - cos²B ]

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Putting this value of second part , we get,

→ cos²A + cos²B + cos²C + 2(cosA * cosB * cosC)

→ cos²A + cos²B + cos²C + [ 1 - cos²C - cos²A - cos²B ]

→ 1 + cos²A - cos²A + cos²B - cos²B + cos²C - cos²C

→ 1 (Ans.)

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