Question

if a<b and c>0 then a-c>b-c where a, b, c and d are real number and c is not equal to 0​

Answer 1

Answer:

Here's a proof by contrapositive. If bc<ac, then 0<ac-bc=(a-b)c. We have two numbers multiplying to equal a positive number, so they are either both positive or both negative. That is:

(a-b>0 and c>0) or (a-b<0 and c<0). Adding b to both sides in the first inequality of each case gives (a>b and c>0) or (a<b and c<0). Examining both cases, it cannot be the case that a>b and c<0. Therefore by contrapositive, a>b and c<0 implies bc>ac.

Answer 2

Answer:

SOLUTION :  

Given : ax² + bx + c = 0 …………(1)  

and - ax² + bx + c = 0…………..(2)  

On comparing the given equation with Ax² + Bx + C = 0    

Let D1 & D2 be the discriminants of the two given equations .  

For eq 1 :    

Here, A = a  , B =  b , C = c  

D(discriminant) = B² – 4AC  

D1 = B² – 4AC = 0  

D1 = (b)² - 4 × a × C  

D1 = b² - 4ac ………….…(3)  

For eq 2 :  

- ax² + bx + c = 0

Here, A = -a  , B =  b , C = c

D(discriminant) = B² – 4AC

D2 = (b)² - 4 × -a × c

D2 = b²  +  4ac…. …………(4)  

Given : Roots are real for both the Given equations i.e D ≥ 0.  

D1 ≥ 0    

b² - 4ac ≥ 0  

[From eq 3]

b²  ≥ 4ac  …………..(5)  

D2 ≥ 0  

b²  +  4ac ≥ 0 …………….(6)

From eq 5 & 6 ,  We proved that at least one of the given equation has real roots.

[Given : a,b,c are real number and ac ≠0]

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