Given that one root of the equation ax^2+bx+c=0 is three times the other, Then 3b^2 = ?

Anuj kumar   10 years AGO

LET TWO ROOT OF EQUATION ARE d band 3d
then according to ques,
sum of root=>
d+3d=-b/a
solving this we get
4d=-b/a..........eqn 1
and product of root
3d^2=c/a..........eqn 2
putting value of d from eqn 1 into 2 we get
3b^2/16a=c
so 3b^2=16ac

Like? Yes (25) | No (7)  

Rats   10 years AGO

Let the roots be x and 3x.
sum of roots=-b/a=4x
squaring both sides
b^2/a^2=16x^2.......(1)
and product of roots=c/a=3x^2.......(2)
Solving (1) and (2), we get
3b^2=16ac.......Ans

Like? Yes (12) | No (2)  

HARJINDER SINGH   10 years AGO

We know that,
sum of roots=(-b/a)
product of roots=c/a
Let the roots are x and 3x.
sum=>
(-b/a)=x+3x=4x
=>x=(-b/4a)...........(1)
product=x.3x=3x^2
x^2=c/3a.................(2)

putting xalue of eq. (1) in eq.(2),we have
b^2/a^2 = 16/3·c/a
Therefore 3b^2=16ac.

Like? Yes (8) | No (1)  

Himanshu Kumar Gupta   10 years AGO

Let first root be 3@
then Other root be @
So sum of roots= -b/a
And products of root = c/a
So 4@=-b/a
@= -b/4a
And 3@^2=c/a
3*b^2/16a^2=c/a
On solving we get 3b^2=16*a*c

Like? Yes (6) | No (1)  

sanjay kumar saha   10 years AGO

sum of the roots=-b/a
4x=-b/a
3x^2=c/a
16x^2=b^2/a^2
b^2=16x^2*a^2
b^2=16*(c/3a)*a^2
3b^2=16ac

Like? Yes (7) | No (2)  

Sandeep R   10 years AGO

the two roots will be:

-b + root(b^2-4ac) and -b - root(b^2-4ac)

so

-b + root(b^2-4ac) = 3(-b - root(b^2-4ac))
=> -b + root(b^2-4ac)= -3b - 3*root(b^2-4ac)
=> 2b = -4(root(b^2-4ac))
=> b = -2(root(b^2-4ac))
on squaring both sides
=> b^2 = 4(b^2-4ac)
=> b^2 = 4b^2 - 16ac
=> 3b^2 = 16ac

Like? Yes (5) | No (1)  

kunwar prabhakar singh   10 years AGO

let first root is x,then other root is 3x
sum of root =x+3X=(-b/a)=4X........(1)
product of root is=x*3X=c/a=3x^2......(2)
squaring eqn 1()
b^2=16*(a^2)*x^2
now multiply both side by 3
3*b^2=16*(a^2)*(3*x^2)
now put the value of 3x^2 from eqn 2nd,ie=c/a
so,3b^2=16a^2*(a/c)=16ac

Like? Yes (5) | No (1)  

FRANKLIN VIJAY   10 years AGO

roots are say x and 3x, their sum will be 4x and their product 3x^2.
b^2/a^2 = 16/3·c/a
3b^2 = 16ac

Like? Yes (4) | No (2)  

devendra kumar gond   10 years AGO

let the roots be x and 3x; then sum will be that is 4x= (-b/a) .......(i)
3x^2= (c/a).......(ii)
squaring equation (i)
we get 16x^2=b^2/a^2;
we need 3b^2 =3*16x^2*a^2
so from equation (ii) 3x^2=c/a
so finally we get 3b^2=16ac

Like? Yes (4) | No (2)  

Aman Kohli   10 years AGO

Two roots are [(-b)+((b^2 - 4*a*c)^(1/2))]/(2*a) ;
and [(-b)-((b^2 - 4*a*c)^(1/2))]/(2*a)

after applying the conditions

[(-b)+((b^2 - 4*a*c)^(1/2))]/(2*a) = 3* [(-b)-((b^2 - 4*a*c)^(1/2))]/(2*a)


on solving
we get

[(-b)+((b^2 - 4*a*c)^(1/2))] / [(-b)-((b^2 - 4*a*c)^(1/2))] = 3/1

applying inverse compendo n dividendo rule

we get

(-b)/((b^2 - 4*a*c)^(1/2)) = 2/1

squaring bth the sides

b^2 = 4*(b^2 -4*a*c)
b^2 = 4(b^2) -16*a*c
3*(b^2) = 16*a*c

Like? Yes (2) | No (1)  

ashiwani kumar singh   10 years AGO

according to shreedracharya formula for quadratic eqution root are
x1=(b+(b^2-4ac))/2a
x2=(b-(b^2-4ac))/2a

x1=3x2

Like? Yes (1) | No   

Shashi Kumar Gupta   10 years AGO

let roots are X & Y....
so,X+Y=-b/a......1
and X*Y-c/a......2
given, X=3Y
by solving eq 1 and eq 2...
we get 3b^2=16ac

Like? Yes (1) | No   

vaibhav rajhns   10 years AGO

let x & 3x be roots
then x +3x =-b/a
x*3*x=c/a
so x=sqrt(c/3*a)
and x=-b/4a
by equating 1&2

so we get 3*b^2=16ac

Like? Yes (1) | No   

Bholanath    10 years AGO

let one root is p
than 2nd root will be 3p
sum of roots =p+3p=b/a
i.e. 4p=b/a
p=b/4a
product of roots=3p*p=c/a
i.e. 3*b/4a*b/4a=c/a
i.e. 3b^2=16ac

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DINKAR BHARTI   10 years AGO

Let root be y
then y+3y=-b/a
so y=-b/4a.....(1)
also y*3y=c/a
3y^2=c/a.......(2)
substituting eq(1) in eq(2)we get
b^2=16ac/3
multiplying both sides by 3 we get
3b^2=16ac
hence opt 7 is correct

Like? Yes (2) | No (1)  

ShAg   10 years AGO

let roots be p and q
we know pq = (c/a) and p+q = (-b/a)
as given q = 3p;
we have; 3p^2 = c/a
and 4p = -b/a
=> b^2 = 16 p^2 a^2
= now putting p^2 = (c/3a)
we have 3b^2 = 16ac

Like? Yes (2) | No (1)  

ABHISHEK KUMAR MISHRA   10 years AGO

let root are r,3r
so r+3r=4r=-b/a....(1) and r*3r=3r^2=c/a...(2)
(2)devided(1) then 3r/4=-c/b...(3)
now from (1) r=-b/4a put it in to (3) then 3b^2=16ac

Like? Yes (1) | No   

rohit kumar   10 years AGO

the sum of the roots is -b/a and their product is c/a.
If the roots are, say, x and 3x, their sum will be 4x and their product 3x^2
So sum squared = 16x^2 = 16/3 times product.
So b^2/a^2 = 16/3*c/a
3b^2 = 16ac

Like? Yes (1) | No   

harikrishna   10 years AGO

suppose x^2-4*x+3 here r1=1, r2=3, a=1, b=-4, c=3
from that 3*-4*-4=16*1*3
like that x*x-8*x+12 r1=2 r2=6 a=1 b=-8 c=12
from that 3*-8*-8 = 16*1*12
so that we can conclude
3*b*b = 16*a*c

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Devendra Marghade   10 years AGO

If x & y are the roots of the equation ax^2+bx+c=0
and given x=3y
Sum of roots=x+y=-b/a, 3y+y=-b/a 4y=-b/a or y=-b/4a ---(i)
Product of roots=xy=c/a, 3y*y=c/a or 3y^2=c/a --(ii)
Substituting value of y from (i) into (ii),
3(-b/4a)^2=c/a so 3b^2=16ac

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Harjit kaur   10 years AGO

1root=p
2root=3p
Now sum of roots. 4p=-b/a
Products of roots. 3p^2=c/a
By solving these we get
3b^2=16ac

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kandula v v sadguru murthy chowdary   10 years AGO

sum of the roots=x+3x=4x=-b/a
product of roots=3x^2=c/a
3b^2=48x^2a^2=16ac

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akshay anand   10 years AGO

x*y=c/a
&x=3y
3y^2=c/a
y=sqrt(c/3a)
putting y in equation we get
b^2=16ac/3
3b^2=16ac

Like? Yes (1) | No   

vigneshwari   10 years AGO

given eqn is ax^2 + bx + c = 0.
=> x^2 + b/a + c/a = 0.
let p and q be two roots of the eqn.
then q = 3p (according to given qstn).
we have x^2 - (p+q)x + pq = 0.
therefore, -(p+q) = b/a and pq = c/a
then, -(p+3p) = b/a and p(3p) = c/a.
-4p = b/a and 3p^2 = c/a
squaring both sides,
16p^2 = (b/a)^2
16 x p^2 x a^2 = b^2
and 3b^2 = 3x 16 x p^2 x a^2
put 3p^2 = c/a in above eqn,we have
3b^2 = 16 x c/a x a^2 = 16ca.

Like? Yes (2) | No (1)  

Yallapu Uday tarun kumar   10 years AGO

take one root as 1/d and other is 3/d
using roots form a eqation and compare with given equation and calculate the value of 3b^2.

Like? Yes (1) | No   

Jasmine Khan   10 years AGO

let x1 and x2 be the root of the given equation,
then,
x1=[-b+sqrt(b^2-4*a*c)]/2*a and x2=[-b-sqrt(b^2-4*a*c)]/2*a
but it is given that the one root is the third times of the another root
i.e. x1=3*x2
[-b+sqrt(b^2-4*a*c)]/2*a=3*[-b-sqrt(b^2-4*a*c)]/2*a
2b/2a=-4sqrt(b^2-4ac)/2a
b=-2sqrt(b^2-4ac)
now squaring both the sides we get,
b^2=4b^2-16ac
3b^2=16ac
hence the answer is option 7.

Like? Yes (1) | No   

Koushik Ghosh   10 years AGO

You know that the sum of the roots is -b/a and their product is c/a.
If the roots are, say, x and 3x, their sum will be 4x and their product 3x^2.
So sum squared = 16x^2 = 16/3 times product.
So b^2/a^2 = 16/3·c/a
Clearing denominators,
3b^2 = 16ac

Like? Yes (1) | No   

SAMRAT CHOWDHURY   10 years AGO

let 2 root be r & 3r
r+3r=-b/a; 4r=-b/a------(1)
r*3r=c/a; 3r^2=c/a------(2)
from eq 1 & 2
3b^2=16ac

Like? Yes (1) | No   

Totan Shit   10 years AGO

roots are r and 3r
r+3r=-(b/a)
so r^2=b^2/(16a^2).....(1)
& 3r^2=c/a
so r^2=c/(3a)..........(2)
solving the two equs
b^2/(16a^2)=c/3a
so 3b^2=16ac(ans)

Like? Yes (1) | No   

lucky raj   10 years AGO

Let two roots of the equation is α & β.

α =(-b+√(b^2-4ac))/2a &

β=(-b-√(b^2-4ac))/2a
given that
α=3β
or,(-b+√(b^2-4ac))/2a
=3((-b-√(b^2-4ac))/2a)
or,-b+√(b^2-4ac)=3(-b-√(b^2-4ac))
or,-b+3b=-(√(b^2-4ac)+3√(b^2-4ac))
or,2b=-4√(b^2-4ac)
or,b=-2√(b^2-4ac)
or,b^2=(-2√(b^2-4ac))^2
[squaring both side]
or,b^2=4(b^2-4ac)
or,b^2=4b^2-16ac
or,4b^2-b^2=16ac
or,3b^2=16ac

Like? Yes (1) | No   

Nirmalram   10 years AGO

one root=3(another root)
=>(-b-sqrt(b^2-4ac))/2a=3((-b+sqrt(b^2-4ac))/2a)
=>(-b-sqrt(b^2-4ac))=3((-b+sqrt(b^2-4ac)))
=>(-b-sqrt(b^2-4ac))=(-3b+3sqrt(b^2-4ac))
=>2b=4sqrt(b^2-4ac)
=>b=2sqrt(b^2-4ac)
=>b/2=sqrt(b^2-4ac)
=>(b^2)/4=b^2-4ac
=>(b^2)=4(b^2)-16ac
=>3(b^2)=16ac

answer:16ac

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Prakash Singh   10 years AGO

we know for a quadratic equn.
ax^2+bx+c having root m,3m
m+3m=-(b/a)
(m*3m)=(c/a)
solving these two we get 3b^2=16ac

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KARIMUL SANFUI   10 years AGO

by using sreedhar acharya we get x1=(-b+(sqrt(b^2-4*a*c)))/2. and x2=(-b-(sqrt(b^2-4*a*c)))/2.
take x1=3*x2 or x2=3*x1.we get 3*b^2=16ac.

Like? Yes (1) | No   

Anantharaman   10 years AGO

let x and y be the roots
given: x=3y
x+y=-b/a
sub x=3y => 3y+y=-b/a
4y=-b/a
sub y=(-b-sqrt(b^2-4ac))/2a then
4 (-b-sqrt(b^2-4ac))/2a = -b/a
2(-b-sqrt(b^2-4ac))/a = -b/a
Canceling denominator a => 2(-b-sqrt(b^2-4ac))=-b
Simplifying => -2b-2*sqrt(b^2-4ac)=-b
-2b+b-2*sqrt(b^2-4ac)=0
-b=2*sqrt(b^2-4ac)
Squaring both sides, we get b^2=4(b^2-4ac)
b^2=4b^2-16ac
3b^2=16ac
Ans: 16ac (Option 7)

Like? Yes (1) | No   

vipul malik   10 years AGO

let the roots of equation be x and y
apply roots x = (-b+(b^2-4ac)^(1/2))/2a
and y = (-b-(b^2-4ac)^(1/2))/2a
now, put x=3y
and get b^2=16ac

Like? Yes (1) | No   

prakash kumar   10 years AGO

let roots are p and 3p
put x=p and 3p,
ap^2+bp+c=0--------------1
9ap^2+3bp+c=0------------2
solve eqn 1 & 2
we get, p = -8c/6b
put value of p in eqn 1
we get, 3b^2=16ac

Like? Yes (1) | No   

RAVI GANGWAR   10 years AGO

You know that the sum of the roots is -b/a and their product is c/a.
If the roots are, say, x and 3x, their sum will be 4x and their product 3x^2.
So sum squared = 16x^2 = 16/3 times product.
So b^2/a^2 = 16/3·c/a
Clearing denominators, 3b^2 = 16ac

Like? Yes (1) | No   

sanjay singh   10 years AGO

let m and 3m be roots
so, m+3m=-b/a (sum of roots)
product of roots m*3m=c/a
eliminating m, we get
3b^2=16ac

Like? Yes (1) | No   

KALIDASS   10 years AGO

the two roots of the given equation is
-b-(b^2-4ac)^1/2,-b+(b^2-4ac)^1/2
now -b-(b^2-4ac)^1/2=3(-b+(b^2-4ac)^1/2
to simplify this we get
b^2=4(b^2-4ac)
3b^2=16ac

Like? Yes (1) | No   

prabhat kumar   10 years AGO

ax^2+bx+c=0
let x and y be two roots
hence y=3x according to question
x+y=-b/c
squaring both side:
(x+y)^2=b^2/a^2
put the value y=3x
x^2=b^2/16a^2--------(1)

x.y=c/a

put y=3x
3x^2=c/a
x^2=c/3a--------(2)
solving (1) and (2)

we get 3b^2=16ac

Like? Yes (1) | No   

J.THARVEES MAIDEEN   10 years AGO

roots of equation are
x1=-b+d/2a; let d=sq root of(b^2-4ac)
x2=-b-d/2a;
given that
x1=3x2
-b+d/2a =3(-b-d/2a)
-b+d =-3b-3d
4d=-2b
take squaring on both side & simplify we get
4d^2=b^2
4(b^2-4ac)=b^2
then 3b^2=16ac

Like? Yes (1) | No   

Raunak MAnchanda   10 years AGO

Applying sum of product and multiplication of product.

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ABHISHEK CHANDEL   10 years AGO

let k,l be roots, so a/t question k=3l,
so 4l=-b/a ...(1),and 3l^2=c/a ...(2)
now 16l^2=(b/a)^2,
put l^2 value from eqn-2,we get 3b^2=16ac

Like? Yes (1) | No   

Grishma Shah   10 years AGO

The 2 roots say A & B of the above equation.
A,B= (-b (+-)rootb^2-4ac)/2a
A=3B
Solving above
Hence 3b^2=16ac

Like? Yes (2) | No (1)  

Bingumalla Abinay   10 years AGO

x 3x be roots
sum=4x and product=3x^2
sum = -b/a and product= c/a
sume^2=16X^2=(b/a)^2
x^2 from product we get c/3a
equating both we get 3b^2=16ac

Like? Yes (2) | No (2)  

Divya Srivastava   10 years AGO

solve taking the values of both roots ans sum & product of roots.

Like? Yes (1) | No (1)  

ravi singh   10 years AGO

answer is 16ac

the root of the given equation is A = [-b+sqrt(b^2-4ac)]/2a ,
B = [-b-sqrt(b^2-4ac)]/2a

now according to the question

A = 3*B

by solving this we get 3b^2=16ac

hence option 7 is right answer

Like? Yes (1) | No (1)  

farjana sayyed   10 years AGO

sum of roots =(-b/a)
product of roots=(c/a)
let the root be
either m,3m (or) -m,-3m (since one root is thrice the other root)
now,
consider the roots -m,-3m
sum of roots=-4m=-b/a......(1)
product of roots=3m*m=c/a....(2)
(2)=> m*m=c/(3a)
squaring on bothside eqn(1)
16m*m=b*b/a*a
16(c/3a)*a*a=b*b
b*b=16ac/3
3b*b=16ac

Like? Yes (1) | No (1)  

rajesh   10 years AGO

we know that the sum of the roots is -b/a and their product is c/a.
If the roots are, say, x and 3x, their sum will be 4x and their product 3x^2.
So sum squared = 16x^2 = 16/3 times product.
So b^2/a^2 = 16/3·c/a
Clearing denominators 3b^2 = 16ac

Like? Yes (1) | No (1)  

Vikram Saini   10 years AGO

ans is 16ac

Like? Yes (1) | No (2)