CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Elimination Method of Finding Solution of a Pair of Linear Equations

54. #160;If...

Question

$54 . I f a + b + c = 5, a^{2} + b^{2} + c^{2} = 12 a n d a^{3} + b^{3} + c^{3} = 25 t h e n t h e v a l u e o f a^{4} + b^{4} + c^{4} i s (a) \frac{251}{6} (b) \frac{253}{6} (c) \frac{255}{6} (d) \frac{257}{6}$

Open in App

Solution

$Given : a + b + c = 5 . . . (A) a^{2} + b^{2} + c^{2} = 12 . . . (B) a^{3} + b^{3} + c^{3} = 25 . . . (C) Now, we take whole square of eq (A), we get {(a + b + c)}^{2} = 25 a^{2} + b^{2} + c^{2} + 2 ab + 2 bc + 2 ca = 25 12 + 2 (ab + bc + ca) = 25 ab + bc + ca = \frac{13}{2} . . . (1) Now, (a + b + c) (a^{2} + b^{2} + c^{2}) = a^{3} + b^{3} + c^{3} + a^{2} b + b^{2} a + b^{2} c + c^{2} b + c^{2} a + a^{2} c . . . (2) And (a + b + c) (ab + bc + ca) = a^{2} b + b^{2} a + b^{2} c + c^{2} b + c^{2} a + a^{2} c + 3 abc So, a^{2} b + b^{2} a + b^{2} c + c^{2} b + c^{2} a + a^{2} c = (a + b + c) (ab + bc + ca) - 3 abc Put this value in eq (2), we get (a + b + c) (a^{2} + b^{2} + c^{2}) = a^{3} + b^{3} + c^{3} + (a + b + c) (ab + bc + ca) - 3 abc a^{3} + b^{3} + c^{3} - 3 abc = (a + b + c) [(a^{2} + b^{2} + c^{2}) - (ab + bc + ca)] Now, we put values from equations A, B, C and 1 and ger 25 - 3 abc = 5 [12 - \frac{13}{2}] 25 - 3 abc = 60 - \frac{65}{2} 3 abc = - \frac{5}{2} abc = - \frac{5}{6} . . . (3) Now we take whole square of eq (2), and get {(a^{2} + b^{2} + c^{2})}^{2} = 12^{2} (a^{4} + b^{4} + c^{4}) + 2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}) = 144 a^{4} + b^{4} + c^{4} = 144 - 2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}) . . . (4) And, {(ab + bc + ca)}^{2} = a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} + 2 (a^{2} bc + {ab}^{2} c + {abc}^{2}) a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = (ab + bc + ca)^{2} - 2 (a^{2} bc + {ab}^{2} c + {abc}^{2}) a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = (ab + bc + ca)^{2} - 2 abc (a + b + c) a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = {(\frac{13}{2})}^{2} - 2 (- \frac{5}{6}) (5) a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = \frac{169}{4} + \frac{50}{6} a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = \frac{607}{12} Put this value in (4), we get a^{4} + b^{4} + c^{4} = 144 - 2 (\frac{607}{12}) a^{4} + b^{4} + c^{4} = \frac{257}{6}$

Suggest Corrections

thumbs-up

0

Similar questions

a + b + c = 3

a^{2} + b^{2} + c^{2} = 6

a^{3} + b^{3} + c^{3} = 8

a^{4} + b^{4} + c^{4} = ?

Q. Lets consider quadratic equation

a x^{2} + b x + c = 0

a, b, c \in R

a \neq 0

. If above equation has roots

α, β

α + β = - \frac{b}{a}, α β = \frac{c}{a}

and the equation can be written as

a x^{2} + b x + c = a (x - α) (x - β)

a_{1}, a_{2}, a_{3}, a_{4}

, ..... are in A.P., then

a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = . . . \neq 0

b_{1}, b_{2}, b_{3}, b_{4}

, ... are in G.P., then

\frac{b_{2}}{b_{1}} = \frac{b_{3}}{b_{2}} = \frac{b_{4}}{b_{3}} =

\neq 1

c_{1}

c_{2}

c_{3}

c_{4}

, ... are in HP, then

\frac{1}{c_{2}} - \frac{1}{c_{1}} = \frac{1}{c_{3}} - \frac{1}{c_{2}} = \frac{1}{c_{4}} - \frac{1}{c_{3}} =

\neq 0

. If the roots of equation

a (b - c) x^{2} + b (c - a) x + c (a - b) = 0

are equal, then

a, b, c

a + b + c = 0

a^{2} + b^{2} + c^{2} = 1

, then the value of

a^{4} + b^{4} + c^{4}

If $a + b + c = 15$ and $a^{2} + b^{2} + c^{2} = 83$ then find the value of $a^{3} + b^{3} + c^{3} - 3 a b c$

a + b + C = 12

a^{2} + b^{2} + c^{2} = 50

a b + b c + a c

a^{3} + b^{3} + c^{3} - 3 a b c

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Algebraic Solution

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Elimination Method of Finding Solution of a Pair of Linear Equations

Standard X Mathematics

Join BYJU'S Learning Program

CrossIcon