Advertisements
Advertisements
Question
If a + b + c = 9 and ab + bc + ca = 23, then a2 + b2 + c2 =
Options
35
58
127
none of these
Advertisements
Solution
We have to find `a^2 + b^2 + c^2`
Given `a+b + c = 9,ab+bc +ca = 23`
Using identity `(a+b+c)^2 = a^2 + b^2 +c^2+2ab + 2bc + 2ca` we get,
`(9)^2 = a^2 +b^2 + c^2+ 2 (ab + bc + ca)`
` 9 xx 9 = a^2 + b^2 + c^2 +2 xx 23`
`81 = a^2 + b^2 + c^2+46`
By transposing +46 to left hand side we get,
`81 - 46 = a^2 +b^2 +c^2`
` 35 = a^2 +b^2 +c^2`
Hence the value of `a^2 +b^2 +c^2` is 35.
APPEARS IN
RELATED QUESTIONS
Expand the following, using suitable identity:
`[1/4a-1/2b+1]^2`
Write the following cube in expanded form:
`[x-2/3y]^3`
Simplify the following products:
`(2x^4 - 4x^2 + 1)(2x^4 - 4x^2 - 1)`
Find the cube of the following binomials expression :
\[\frac{1}{x} + \frac{y}{3}\]
Find the cube of the following binomials expression :
\[2x + \frac{3}{x}\]
If 2x+3y = 13 and xy = 6, find the value of 8x3 + 27y3
Find the value of 27x3 + 8y3, if 3x + 2y = 20 and xy = \[\frac{14}{9}\]
If a + b = 10 and ab = 16, find the value of a2 − ab + b2 and a2 + ab + b2
If a + b = 8 and ab = 6, find the value of a3 + b3
If \[x + \frac{1}{x} = 3\] then find the value of \[x^6 + \frac{1}{x^6}\].
If \[x + \frac{1}{x} = 2\], then \[x^3 + \frac{1}{x^3} =\]
\[\frac{( a^2 - b^2 )^3 + ( b^2 - c^2 )^3 + ( c^2 - a^2 )^3}{(a - b )^3 + (b - c )^3 + (c - a )^3} =\]
Evaluate: (6 − 5xy) (6 + 5xy)
If x + y = 9, xy = 20
find: x - y
If `x + (1)/x = 3`; find `x^4 + (1)/x^4`
If x + y + z = p and xy + yz + zx = q; find x2 + y2 + z2.
If `"a"^2 + (1)/"a"^2 = 14`; find the value of `"a" + (1)/"a"`
If `"r" - (1)/"r" = 4`; find: `"r"^2 + (1)/"r"^2`
Simplify:
(x + y - z)2 + (x - y + z)2
Factorise the following:
9x2 + 4y2 + 16z2 + 12xy – 16yz – 24xz
