Algebra _ Undetermined Coefficients, Exercises of Mathematics

Solutions to five algebraic problems involving undetermined coefficients. The problems require finding the values of variables given an equation. The solutions involve algebraic manipulation and substitution. useful for students studying algebra and preparing for exams or assignments on undetermined coefficients.

Typology: Exercises

Pre 2010

Available from 03/01/2022

SKarthigeyan
SKarthigeyan 🇮🇳

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Algebra
Undetermined coefficients
1. Find a, b, c and d given ax3 + bx2 + cx + d = 7x3 10x2 3x 12.
Solution: ax3 + bx2 + cx + d = 7x3 10x2 3x 12.
a = 7, b = 10 , c = 3 , d = 12
2. If ( x + m)2 x2 + px + 9, Find m and p.
Solution:
(x + m) 2 x2 + px + 9
x2 + 2mx + m2 = x2 + px + 9
m2 = 9
 m = 3
2m = p
p = 6
3. Given (2A +B) x + (A + B) 11x + 7, find the values of A and B.
Solution:
(2A + B)x + (A + B) 11x + 7
2A + B = 11 -----(1)
A + B = 7 -----(2)
Subtract, A = 4
(2) B = 3
pf3
pf4

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Algebra

Undetermined coefficients

1. Find a , b, c and d given ax^3 + b x^2 + c x + d = 7 x^3 – 10 x^2 – 3 x – 12.

Solution: ax^3 + b x^2 + c x + d = 7 x^3 – 10 x^2 – 3 x – 12.

 a = 7, b =  10 , c =  3 , d =  12

2. If ( x + m )^2  x^2 + px + 9, Find m and p****.

Solution: ( x + m ) 2  x^2 + px + 9 x^2 + 2 mx + m^2 = x^2 + px + 9  m^2 = 9

 m =  3

2 m = p

 p =  6

3. Given (2A +B) x + (A + B)11 x + 7, find the values of A and B.

Solution: (2A + B) x + (A + B)  11 x + 7 2A + B = 11 -----(1) A + B = 7 -----(2) Subtract, A = 4 (2)B = 3

4. If M ( x + 3) + N ( x – 2)8 x + 9, find the values of M and N.

Solution: M ( x + 3) + N ( x – 2) = 8 x + 9 M x + 3M + N x – 2N = 8 x + 9 (M + N) x + (3M – 2N) = 8 x + 9

 M + N = 8 ---- (1)

3M – 2N = 9 ---- (2)

(1) x 2  2M + 2N = 16 3 M – 2N = 9 5M = 25  M = 5 (1)N = 3

5. If P( x + 1) ( x + 2) + Q ( x – 1)x^2 + 6 x – 1, find the values of P and Q.

Solution: Put x = 1 P(2) (3) = (1)^2 + 6 (1) – 1 6 P = 6  P = 1 Put x =  1 Q (2) = (1)^2 + 6 (1) – 1 2 Q =  6 Q = 3

8. Given A ( x – 2)^3 + B ( x – 2)^2 + C ( x – 2) + D = 2 x^3 + 8 x^2 + 22 x + 22 find the values of A, B, C and D. Solution: x = 2 D = 2 (8) + 8 (4) + 22 (2) + 22 = 16 + 32 + 44 + 22 = 114 Equate the coefficient of x^3 A = 2 Equate the coefficient of x^2 A (6) + B = 8 B = 20 Put x = 0 A (8) + B (4) + C (2) + D = 22  16 + 80 – 2C + 114 = 22  2C =  156 C = 78

 A = 2, B = 20, C = 78, D = 114.