Question

prove that the value of the expression (a-1)(a-3)(a-4)(a-6)+10 is positive for every a e r

Answer 1

Answer:

refer to solution

Step-by-step explanation:

well if you take any value more than 6 it's positive clearly

if you take 1 3 4 or 6 it is 10

now if you take 2 or 5 you get 2

if you take 0 or any value below it then it comes positive as the product becomes positive

Answer 2

SOLUTION

TO PROVE

The value of the expression (a-1)(a-3)(a-4)(a-6) + 10 is positive for every a ∈ R

EVALUATION

Here the given expression is

$\sf (a - 1)(a - 3)(a - 4)(a - 6) + 10$

We simplify the expression as below

$\sf (a - 1)(a - 3)(a - 4)(a - 6) + 10$

$\sf = (a - 1)(a - 6)(a - 3)(a - 4) + 10$

$\sf = ( {a}^{2} - 6a - a + 6)( {a}^{2} - 3a - 4a + 12) + 10$

$\sf = ( {a}^{2} -7 a + 6)( {a}^{2} - 7a + 12) + 10$

Let x = a² - 7a

Then given expression

$\sf = (x+ 6)( x + 12) + 10$

$\sf = {x}^{2} + 6x + 12x + 72 + 10$

$\sf = {x}^{2} + 18x + 82$

$\sf = {x}^{2} + 18x + 81 + 1$

$\sf = {(x + 9)}^{2} + 1$

$\sf = {( {a}^{2} - 7a + 9)}^{2} + 1$

Which is positive for all real values of a

Hence proved

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