Question

hiii guys ☺☺☺
If the roots are the equation is ax^2+2bx+c =0 and bx^2-2√acx + b = 0 are simultaneously real , then proof that [b^2-ac ] ..plzz solve .........

Answer 1

Hello! sanskriti!
#ur Ans
________________

Given that :
Root of both the equation are real
so,

1st equation :
------------------

➡
$a {x}^{2} + 2bx + c = 0 \\ \\ here \: descriminant \: so$
➡ D ≥ 0

$= > (2b {)}^{2} - 4(ac) \geqslant 0 \\ \\ = > 4 {b}^{2} - 4ac \geqslant 0 \\ \\ = > 4 {b}^{2} \geqslant 4ac \\ \\ = > {b}^{2} \geqslant ac............(1) \\ \\ now \: 2nd \: equation \\ \\ = > b {x}^{2} - 2 \sqrt{(ac)} \: x + b = 0 \\ \\ = > here \: decriminant \\ so$
➡ D ≥ 0
$= >( 2 \sqrt{ac)} { \: ) }^{2} - 4 ({b}^{2} ) \geqslant 0 \\ \\ = > 4ac - 4 {b}^{2} \geqslant 0 \\ \\ = > ac \geqslant {b}^{2} .........(2) \\ \: the \: result \: of \: \: eq \: 1 \: and \: eq \: 2 \: are \: \\ \: simultaneousely \: \: possible \\ \: only \: one \: case \: when \\ \\ = > {b}^{2} = ac \\ \\ or \\ \\ = > b {}^{2} - 4ac \: \\ hence \: proved$

☺✌:-)

Answer 2

SOLUTION :  

Given : ax² + 2bx + c = 0 …………(1)

and bx² - 2√acx + b = 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 =  2b , C = c

D(discriminant) = B² – 4AC

D1 = (2b)² - 4 × a × C

D1 = 4b² - 4ac ………(3)

For eq 2 :  

bx² - 2√acx + b = 0

Here, A = b  , B =  - 2√ac, C = b

D(discriminant) = B² – 4AC

D2 = (- 2√ac)² - 4 × b × b  

D2 = 4ac - 4b² …………(4)

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

D1 ≥ 0  

4b² - 4ac ≥ 0  

[From eq 3]

4b²  ≥ 4ac  

b² ≥ ac  ………….(5)

D2 ≥ 0

4ac - 4b²  ≥ 0

4ac  ≥ 4b²

ac  ≥ b² …………(6)

From eq 5 & 6 ,  

b² = ac  

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