Advertisements
Advertisements
Question
Simplify the following products:
`(x/2 - 2/5)(2/5 - x/2) - x^2 + 2x`
Solution
`(x/2 - 2/5)(2/5 - x/2) - x^2 + 2x`
On rearranging we get,
⇒ `(x/2 - 2/5)[-(x/2 - 2/5)] - x^2 + 2x`
⇒ `- (x/2 - 2/5)^2 - x^2 + 2x`
We shall use the identity (x − y)2 = x2 − 2xy + y2
By substituting `x = x/2, y = 2/5`
⇒ `- [(x/2)^2 - 2(x/2)(2/5) + (2/5)^2] - x^2 + 2x`
⇒ `- [x^2/4 - (2x)/5 + 4/25] - x^2 + 2x`
⇒ `- [x^2/4 - (2x)/5 + 4/25] - x^2 + 2x`
⇒ `- x^2/4 + (2x)/5 - 4/25 - x^2 + 2x`
⇒ `- x^2/4 - x^2 - 4/25 + (2x)/5 + 2x`
⇒ `[- x^2/4 - x^2] - 4/25 + [(2x)/5 + 2x]`
⇒ `[- x^2/4 - x^2] - 4/25 + [(2x)/5 + 2x]`
⇒ `[- x^2/4 - (4x^2)/4] - 4/25 + [(2x)/5 + (10x)/5]`
⇒ `[(- x^2 - 4x^2)/4] - 4/25 + [(2x + 10x)/5]`
⇒ `(- 5x^2)/4 - 4/25 + (12x)/5`
Hence, the value of `(- 5x^2)/4 - 4/25 + (12x)/5`.
APPEARS IN
RELATED QUESTIONS
Expand the following, using suitable identity:
(3a – 7b – c)2
Verify:
x3 + y3 = (x + y) (x2 – xy + y2)
Evaluate the following using identities:
(1.5x2 − 0.3y2) (1.5x2 + 0.3y2)
Write in the expanded form: (ab + bc + ca)2
If 3x − 2y = 11 and xy = 12, find the value of 27x3 − 8y3
Simplify of the following:
\[\left( x + \frac{2}{x} \right)^3 + \left( x - \frac{2}{x} \right)^3\]
If x = 3 and y = − 1, find the values of the following using in identify:
\[\left( \frac{x}{y} - \frac{y}{3} \right) \frac{x^2}{16} + \frac{xy}{12} + \frac{y^2}{9}\]
Find the following product:
(3x − 4y + 5z) (9x2 +16y2 + 25z2 + 12xy −15zx + 20yz)
Use the direct method to evaluate the following products :
(y + 5)(y – 3)
Use the direct method to evaluate :
(ab+x2) (ab−x2)
Use the direct method to evaluate :
`(3/5"a"+1/2)(3/5"a"-1/2)`
Expand the following:
(x - 5) (x - 4)
Find the squares of the following:
9m - 2n
Find the squares of the following:
3p - 4q2
Evaluate, using (a + b)(a - b)= a2 - b2.
15.9 x 16.1
If x + y + z = 12 and xy + yz + zx = 27; find x2 + y2 + z2.
If `"p" + (1)/"p" = 6`; find : `"p"^2 + (1)/"p"^2`
Simplify:
(4x + 5y)2 + (4x - 5y)2
Find the following product:
`(x/2 + 2y)(x^2/4 - xy + 4y^2)`
If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 – 3abc = – 25.
