Advertisements
Advertisements
प्रश्न
If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 + c2.
Advertisements
उत्तर
Given, a + b + c = 9 and ab + bc + ca = 26 ...(i)
Now, a + b + c = 9
On squaring sides, we get
(a + b + c)2 = (9)2
⇒ a2 + b2 + c2 + 2ab + bc + ca = 81 ...[Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
⇒ a2 + b2 + c2(ab + bc + ca) = 81
⇒ a2 + b2 + c2 + 2(26) = 81 ...[From equation (i)]
⇒ a2 + b2 + c2 = 81 – 52 = 29
APPEARS IN
संबंधित प्रश्न
Expand the following, using suitable identity:
(–2x + 5y – 3z)2
Write the following cube in expanded form:
(2a – 3b)3
Factorise the following:
27 – 125a3 – 135a + 225a2
Verify:
x3 + y3 = (x + y) (x2 – xy + y2)
Factorise:
27x3 + y3 + z3 – 9xyz
Without actually calculating the cubes, find the value of the following:
(–12)3 + (7)3 + (5)3
Simplify the following products:
`(x/2 - 2/5)(2/5 - x/2) - x^2 + 2x`
Find the value of 4x2 + y2 + 25z2 + 4xy − 10yz − 20zx when x = 4, y = 3 and z = 2.
Find the following product:
\[\left( 3 + \frac{5}{x} \right) \left( 9 - \frac{15}{x} + \frac{25}{x^2} \right)\]
If x = 3 and y = − 1, find the values of the following using in identify:
(9y2 − 4x2) (81y4 +36x2y2 + 16x4)
If a - b = 0.9 and ab = 0.36; find:
(i) a + b
(ii) a2 - b2.
If a + `1/a`= 6 and a ≠ 0 find :
(i) `a - 1/a (ii) a^2 - 1/a^2`
If a2 - 3a + 1 = 0, and a ≠ 0; find:
- `a + 1/a`
- `a^2 + 1/a^2`
Evaluate: (5xy − 7) (7xy + 9)
Expand the following:
(2x - 5) (2x + 5) (2x- 3)
If `"a" + 1/"a" = 6;`find `"a"^2 - 1/"a"^2`
If `"a"^2 - 7"a" + 1` = 0 and a = ≠ 0, find :
`"a"^2 + (1)/"a"^2`
Simplify:
`(x - 1/x)(x^2 + 1 + 1/x^2)`
Using suitable identity, evaluate the following:
1033
Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (–z + x – 2y).
