Advertisements
Advertisements
Question
If a − b = 5 and ab = 12, find the value of a2 + b2
Solution
We have to find the value `a^2 +b^2`
Given a-b = 5, ab = 12
Using identity `(a - b)^2 = a^2 - 2ab +b^2`
By substituting the value of a-b = 5 ,ab = 12 we get ,
`(5)^2 = a^2 +b^2 - 2 xx 12`
`5 xx 5 = a^2 +b^2 - 2 xx 12`
`25 = a^2 +b^2 -24`
By transposing – 24 to left hand side we get
`25 + 24 = a^2 +b^2`
`49 = a^2 +b^2`
Hence the value of `a^2 +b^2` is 49.
APPEARS IN
RELATED QUESTIONS
Factorise:
`2x^2 + y^2 + 8z^2 - 2sqrt2xy + 4sqrt2yz - 8xz`
Verify that `x^3+y^3+z^3-3xyz=1/2(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]`
Write in the expanded form (a2 + b2 + c2 )2
If \[x - \frac{1}{x} = - 1\] find the value of \[x^2 + \frac{1}{x^2}\]
If x = 3 and y = − 1, find the values of the following using in identify:
\[\left( \frac{3}{x} - \frac{x}{3} \right) \left( \frac{x^2}{9} + \frac{9}{x^2} + 1 \right)\]
If x = 3 and y = − 1, find the values of the following using in identify:
\[\left( \frac{5}{x} + 5x \right)\] \[\left( \frac{25}{x^2} - 25 + 25 x^2 \right)\]
If a + b = 10 and ab = 16, find the value of a2 − ab + b2 and a2 + ab + b2
If \[x - \frac{1}{x} = \frac{1}{2}\],then write the value of \[4 x^2 + \frac{4}{x^2}\]
If a + b + c = 9 and ab + bc + ca = 23, then a2 + b2 + c2 =
If a + b + c = 0, then \[\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} =\]
Use identities to evaluate : (998)2
Find the squares of the following:
(2a + 3b - 4c)
Simplify by using formula :
(a + b - c) (a - b + c)
Evaluate, using (a + b)(a - b)= a2 - b2.
15.9 x 16.1
If x + y = 9, xy = 20
find: x - y
If `x + (1)/x = 3`; find `x^2 + (1)/x^2`
Simplify:
(3a + 2b - c)(9a2 + 4b2 + c2 - 6ab + 2bc +3ca)
If a + b + c = 0, then a3 + b3 + c3 is equal to ______.
Expand the following:
`(4 - 1/(3x))^3`
Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (–z + x – 2y).