Question

if a,b,and c real numbers such that ac is not equal to 0,then show that at least one of the equations ax^2+bx+c and -a^2+bx+c =0. has equal zeros

Answer 1

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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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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