Question

If 0>a>b>c, and a^2 + bc = 20, b^2 + ca = 20 and c^2 + ab = 30, what is the value of a, b and c?

Answer 1

Heya !!

Here is your solution :

Given,

⇒ 0 > a > b > c

And,

⇒ a² + bc = 20 --------- ( 1 )

⇒ b² + ca = 20 -------- ( 2 )

⇒ c² + ab = 30 --------- ( 3 )

Adding ( 1 ) , ( 2 ) and ( 3 ),

⇒ a² + b² + c² + bc + ca + ab = 20 + 20 + 30

•°• a² + b² + c² + ab + bc + ca = 70 ------- ( 4 )

From ( 1 ) and ( 2 ) we get,

⇒ a² + bc = b² + ca

⇒ a² - b² = ca - bc

⇒ ( a + b ) ( a - b ) = c ( a - b )

•°• ( a + b ) = c ---------- ( 5 )

Adding ( 1 ) and ( 2 ),

⇒ a² + bc + b² + ca = 20 + 20

⇒ a² + b² + bc + ca = 40

⇒ a² + b² + c ( a + b ) = 40

Substitute the value of ( 6 ),

⇒ a² + b² + c × c = 40

⇒ a² + b² + c² = 40 -------- ( 6 )

Substitute the value of ( 6 ) in ( 4 ),

⇒ a² + b² + c² + ab + bc + ca = 70

⇒ 40 + ab + bc + ca = 70

⇒ ab + bc + ca = 70 - 40

⇒ ab + bc + ca = 30

Multiplying both sides by 2,

⇒ 2( ab + bc + ca ) = 2 × 30

•°• 2ab + 2bc + 2ca = 60 -------- ( 7 )

Adding ( 6 ) and ( 7 ),

⇒ a² + b² + c² + 2ab + 2bc + 2ca = 40 + 60

⇒ ( a + b + c )² = 100

⇒( a + b + c ) = √100

•°• ( a + b + c ) = -10 -------( 8 )
[ Neglecting +ve value ]

Subtracting ( 8 ) from ( 5 ),

⇒( a + b ) - ( a + b + c ) = c - ( -10 )

⇒ a + b - a - b - c = c + 10

⇒ -c = c + 10

⇒ -c - c = 10

⇒ -2c = 10

⇒ c = 10 ÷ ( -2 )

•°• c = -5

Substitute the value of 'c' in ( 6 ),

⇒ a² + b² + c² = 40

⇒ a² + b² + ( -5 )² = 40

⇒ a² + b² + 25 = 40

⇒ a² + b² = 40 - 25

•°• a² + b² = 15 ---------- ( 9 )

Substitute the value of 'c' in ( 5 ),

⇒ ( a + b ) = c

⇒ a + b = ( -5 ) --------( 10 )

Squaring both sides,

⇒ ( a + b )² = ( -5 )²

•°• a² + b² + 2ab = 25 --------- ( 11 )

Substitute the value of ( 9 ) in ( 11 ),

⇒ a² + b² + 2ab = 25

⇒ 15 + 2ab = 25

⇒ 2ab = 25 - 15

•°• 2ab = 10 --------- ( 12 )

Subtact ( 12 ) from ( 9 ),

⇒ a² + b² - 2ab = 15 - 10

⇒( a - b )² = 5

⇒ ( a - b ) = √5 --------- ( 12 )
[ Neglecting ( -ve ) value ]

Adding ( 12 ) and ( 10 ),

⇒ a + b + a - b = -5 + √5

⇒ 2a = √5 - 5

$⇒ \: a \: = \: \frac{ \sqrt{5} \: - \: 5 }{2}$

Subtracting ( 12 ) from ( 10 ),

⇒ ( a + b ) - ( a - b ) = -5 - √5

⇒ a + b - a + b = -5 - √5

⇒ 2b = -5 - √5

$⇒ \: b \: = \: \frac{ - 5 \: - \sqrt{5} }{2}$

Hence,

$⇒ \: a \: = \: \frac{ \sqrt{5} \: - \: 5 }{2}$

$⇒ \: b \: = \: \frac{ - 5 \: - \sqrt{5} }{2}$

$⇒ \: c \: = \: - 5$

Hope it helps !!

Answer 2

Thanks for a fantabalous question.


Solution contains all steps.



On finding any mistake, comment immediately.