Question

If one the roots
the rools of the equation
x² +ax+3=0
is 3
and
one of the roots of the equation
x² +ax+b=0 is three times of the
other roots. then b=?
if anyone answers this question
I will mark their all answers as brainlyist
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Answer 1

Answer:

b=3

Step-by-step explanation:

since one of the roots of x^2+ax+3=0 is 3

the product of roots=constant term÷coefficient of x^2=3÷1=3

the other root=3/3=1

sum of roots=(-coefficient of x)÷coefficient of x^2

3+1=(-a)/1

a=(-4)

put this value of a in second equation it becomes

x^2-4x+b

let smaller root of the equation be x

then the other root becomes 3x

x+3x=-(-4)/1=4

x=1

3x=3

3×1=b/1

b=3

please mark my answer as brainliest

Answer 2

AnswEr:-

Given , one of the roots of x²+ax+3=0 is 3

→ Product of roots=constant term / coefficient of x²

    = 3 / 1 = 1.

The other root  will be  3/3=1

→sum of roots=(-coefficient of x)/ coefficient of x²

⇒ 3+1=(-a)/1

⇒a=(-4)

Substitute  this value of a in second equation :

⇒ x²-4x+b

let smaller root of the equation be x

then the other root  will be  3x

⇒ x+3x=-(-4)/1=4

⇒ x=1

⇒3x=3

⇒3×1=b/1

⇒ b=3

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