Saidinesh   10 years AGO

2x^2+kx-5=0.........(1)
x^2-3x-4=0..........(2)
solving equation 2
x^2-3x+x-4=0;
(x+1)(x-4)=0;
x=-1;x=4;
put x value in first equation
x=-1;
2-k-5=0;
k=-3;
x=4;
32+4k-5=0
k=-27/4

Like? Yes (16) | No (5)  

Dr.Dipin Singh   10 years AGO

x^2 - 3x - 4 = 0
(x-4)(x+1)=0 means two roots are 4, and -1.
when common root is -1,
Value of k in 2x^2+kx-5 = 0 is -3.
when common root is 4,
value of k in 2x^2+kx-5 = 0 is -27/4.

Like? Yes (13) | No (7)  

Mahalakshmi   10 years AGO

By solving equation x^2-3x-4=0 will get x=4,-1
by substituting x value in equation 2x^2+kx-5 =0 will get k= -3,-27/4

Like? Yes (5) | No (1)  

abhilasha   10 years AGO

Answer -3, -27/4
on keeping the values -3, -27/4 we get the roots -1,4 respectively.And from the second equation we got the roots 4,-1 so the roots are matching and thus this is the answer.

Like? Yes (2) | No   

S.SHAFI   10 years AGO

The roots of the equation x^2-3-4=0 are 4,-1...
It can be calculated by using

x = {=-b+-sqrt(b^2-4*a*c)}/2a

Substitute k=-3 and -27/4 in the equation 2x^2+kx-5=0 to have one root in common.

When k=-3, the roots are 5/2, -1

When k=-27/4, the roots are -5/8, 4

So, ans is option 3...

Like? Yes (4) | No (3)  

T G   10 years AGO

ans is 3
the root of the 2nd equation is 4 and -1.
now put the value of x in the 1 st equation respectively....get the result
-3 and -27/4

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R.RAVIKUMAR   10 years AGO

for second eqn we get roots as 4 and -1.
by substituting -3,-27/4 in first eqn, we get roots as -1 and 4 respectively...

Like? Yes (2) | No (2)  

K Venkata Subba Rao   10 years AGO

2x^2+kx-5 =0
-2*(x^2-3x-4 )=0
(+)---------------
x= -3/(k+3)
(sub x in eq)
u'll get k=-3 or -27/4

Like? Yes (1) | No (1)  

SID PRASANTH KUMAR   10 years AGO

%alpha ={ca1 -ac1 } over {ab1 - ba1 }={bc1 - cb1 } over {ca1 - ac1 }

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Varunraj Reddy   10 years AGO

From the equation 2 we get the roots are 4,-1

For the value of k=-3,-27/4 we get the same roots
from the equation 1

So option 3) is answer.

Like? Yes (1) | No (1)  

IFFAT IRAM   10 years AGO

let 'a' be the common root
2a^2+ka-5=0........(1)
a^2-3a-4=0.........(2)
by cross multiplication,


a^2/(-4k-15)=a/(8-5)=1/(-6-k)
now,
a^2=(4k+15)/(6+k)
a=-3/(6+k)

again,

9/(6+k)^2=(4k+15)/(6+k)
(4k+15)(6+k)=9
4k^2+39k+81=0
(4k+27)(k+3)=0
k=-27/4 or -3

Like? Yes (2) | No (2)  

j.b.manasa   10 years AGO

x^2-3x-4=0
=(x-4)*(x+1)=0
=x=-1,4.
Common roots means when these x values obtained are substituted in the equation1 then it must be 0
2(-1)^2+k(-1)-5=0 and 2(4)^2+K(4)-5=0
=2-k-5=0 and 32+4k-5=0
k=-3,-27/4

Like? Yes (1) | No (1)  

kothakota sai priya   10 years AGO

x^2-3x-4=0 solve this equation we get x values as
x=4 and x=-1
substitute these x values in other equation i.e
for x=4

2(4)^2+k(4)-5=0
32+4k-5=0
k=-27/4

for x=-1
2(-1)^2+k(-1)-5=0
2-k-5=0
k=-3

Like? Yes (1) | No (1)  

MP    10 years AGO

two roots of x^2 - 3x - 4 = 0 are 4 and -1.
consider one of these ( one by one) as common root with 2x^2+kx-5=0,
we get k = -3,-27/4

Like? Yes (3) | No (3)  

Sourav Das   10 years AGO

2nd eqn has two roots x=4,-1.
put these value in the 1st eqn and solve for the value of k.
k=-27/4 for x=4
k=-3 for x=-1.

Like? Yes (2) | No (2)  

Nitesh Jadhav   10 years AGO

solve using [-b (+/-)root(b^2-4ac)]/2a formula

Like? Yes (1) | No (1)  

sai sarma   10 years AGO

x^2-3x-4 can be divided as
x^2-4x+x-4
=>x(x-4)+1(x-4)
=>x=4,x=-1

we have a condition may be one root matches
take x=4
then sub in eqn1 then k=-27/4
take x=-1 and k=-3

so option (3) is correct

Like? Yes (2) | No (2)  

hema   10 years AGO

roots of second equations are 4 and -1.
substitute in 1st equ

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susmit   10 years AGO

By solving 2nd equation we get x = -1 or x = 4
Substituting both the values in the first equztion, we get k = -3 or k = -27/4

Like? Yes (2) | No (2)  

Udhaya kumar   10 years AGO

For solving 2nd eqn we get
x=4,-1
sub x in first eqn we get
k = -3,-6.75(-27/4).......

Like? Yes (1) | No (1)  

Sandeep Kumar   10 years AGO

x^2-3x-4 =0
it gives.....
x= 4,-1

putting these two values in eq 1 we get,
k= -27/4 , -3

Like? Yes (1) | No (1)  

Vikram Saini   10 years AGO

x^2-3x-4=0
solve it nd get x=4,-1
nd put the valu in othr eqn
nd get k=-27/4,-3
so option 3 is corrct answr.
thnxxxx

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Jigyasa Kumar   10 years AGO

Answer -3, -27/4
on keeping the values -3, -27/4 we get the roots -1,4 respectively.And from the second equation we got the roots 4,-1 so the roots are matching and thus this is the answer.

Like? Yes (1) | No (2)  

Aarush Singh   10 years AGO

Answer -3, -27/4
on keeping the values -3, -27/4 we get the roots -1,4 respectively.And from the second equation we got the roots 4,-1 so the roots are matching and thus this is the answer.

Like? Yes (1) | No (2)  

ASHISH MITTAL   10 years AGO

x^2 - 3x - 4 = 0
root of this equation after solving
x=-1,4
And after putting value of x in 2x^2+kx-5 = 0
we get x=-3, -27/4

Like? Yes (1) | No (2)  

Krishna Chaitanya   10 years AGO

Given equation: 2x^2+kx-5 = 0 & x^2 - 3x - 4 = 0 may have one root common.
On solving x^2 - 3x - 4 = 0 we get roots -1,4
By substituting x values in 2x^2+kx-5 = 0 we get k values which are -3,-27/4

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Devendra Marghade   10 years AGO

2x^2+kx-5 = 0 ---(i) &
x^2-3x-4 =0 ---(ii)
Solving (ii),x^2+x-4x-4=0, (x+1)(x-4)=0,x=-1,4
For value of k,
Substituting x=-1 in eqn.(i),2(-1)^2+k(-1)-5=0
k=-3 &
with x=4 in eqn.(i),2(4)^2+k(4)-5=0,
k=-27/4

Like? Yes (1) | No (2)  

Srinath   10 years AGO

solve x^2 - 3x - 4 = 0, X=-1,4
pui it on 2x^2+kx-5 = 0 ,
we get k=-3,-27/4 respectively...

Like? Yes (1) | No (2)