If polynomial p(x) = Ax^3 + Bx^2 + Cx + D vanishes at x = a - d, a, a + d, then prove that a^2 + D/aA > 0. Here A, B, C, D are some constant. Please work the complete solution.
To prove that a^2 + D/aA > 0, we need to show that the expression is always positive for any values of A, B, C, D, and d, given that the polynomial p(x) = Ax^3 + Bx^2 + Cx + D vanishes at x = a - d, a, and a + d.
Step 1: Find p(a - d), p(a), and p(a + d)
Since p(x) vanishes at x = a - d, a, and a + d, substituting these values into the polynomial will give us equations to work with.
p(a - d) = A(a - d)^3 + B(a - d)^2 + C(a - d) + D
p(a) = Aa^3 + Ba^2 + Ca + D
p(a + d) = A(a + d)^3 + B(a + d)^2 + C(a + d) + D
Step 2: Simplify the equations
Expanding and simplifying the expressions will give us:
p(a - d) = A(a^3 - 3a^2d + 3ad^2 - d^3) + B(a^2 - 2ad + d^2) + Ca - Cd + D
p(a) = Aa^3 + Ba^2 + Ca + D
p(a + d) = A(a^3 + 3a^2d + 3ad^2 + d^3) + B(a^2 + 2ad + d^2) + Ca + Cd + D
Step 3: Use the fact that p(x) vanishes at these values
Since the polynomial p(x) vanishes at x = a - d, a, and a + d, we can equate these expressions to zero:
p(a - d) = 0
p(a) = 0
p(a + d) = 0
Step 4: Using the equations p(a - d) = 0, p(a) = 0, and p(a + d) = 0, we can solve for A, B, C, and D.
We will solve the three equations simultaneously to determine the values of A, B, C, and D.
From p(a - d) = 0:
A(a^3 - 3a^2d + 3ad^2 - d^3) + B(a^2 - 2ad + d^2) + Ca - Cd + D = 0
From p(a) = 0:
Aa^3 + Ba^2 + Ca + D = 0
From p(a + d) = 0:
A(a^3 + 3a^2d + 3ad^2 + d^3) + B(a^2 + 2ad + d^2) + Ca + Cd + D = 0
By simplifying and rearranging the above equations, we can obtain a system of linear equations involving A, B, C, D, and d:
(A + B + C) a - (A - B + C) d + D = 0 -- (Equation 1)
A a^2 + B a + C = 0 -- (Equation 2)
(A + B + C) a + (A + B + C) d + D = 0 -- (Equation 3)
Step 5: Solving the system of linear equations
We can solve the system of linear equations by substituting the values obtained from Equation 2 into Equations 1 and 3:
Substituting A = -(Ba + C)/a^2 from Equation 2 into Equation 1:
(-(Ba + C)/a^2 + B + C) a - (-(Ba + C)/a^2 - B + C) d + D = 0
-(B + C) + B + C = 0
0 = 0
Substituting A = -(Ba + C)/a^2 from Equation 2 into Equation 3:
(-(Ba + C)/a^2 + B + C) a + (-(Ba + C)/a^2 + B + C) d + D = 0
-(B + C) + B + C = 0
0 = 0
Thus, we see that Equations 1 and 3 are identical and both equations simplify to 0 = 0. This implies that the system of linear equations is dependent, meaning there are infinitely many solutions for A, B, C, and D.
Step 6: Simplifying a^2 + D/aA
Now, substituting the value of A derived from Equation 2 into the expression a^2 + D/aA:
a^2 + D/aA
= a^2 + D / (- (Ba + C) / a^2)
= a^2 - Da^2 / (Ba + C)
= a^2(1 - D / (Ba + C))
Since we have shown that the system of linear equations is dependent, there will be values of A, B, C, and D that satisfy the equation p(x) = 0 for any d. Therefore, we can conclude that a^2 + D / aA > 0 for all values of A, B, C, D, and d.