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If one root of $x^{2} + px + 12 = 0$ is $4$, while the equation $x^{2} + px + q = 0$ has equal roots, then the value of $q$ is:

1. $49/4$
2. $4/49$
3. $4$
4. $\frac{1}{4}$
asked | 59 views

We know :

In quadratic equation :  ax2 + bx +  c = 0

a) Sum of roots  =  (-b/a)

b) Product of roots  =  (c/a)

Hence for the equation : x2 +px + 12 = 0 , we are given one root as 4 and c = 12 which is product of the two roots.

Thus the other root  =  12 / 4  =  3

==> Sum of roots    =  4 + 3  =  7    ==> -p  =  7  ==> p = -7

Now for the other given equation :   x2 +px + q = 0  , which can be now written as : x2  - 7x + q = 0

It is given that this equation has equal roots .

We know :

For equal roots , we have discriminant  " b2 - 4ac = 0"

So here we have :

49 - 4q = 0

==>  q         =  49 / 4

Hence A) should be the correct option.

answered by (1.4k points) 1 3 12
selected
+1 vote

If one root of x^2+px+12=0 is 4 and x^2+px+q=0 has equal roots

answered by (552 points) 1 4 22

if one root is given then we the root must satisfy its equation so 16 + 4p +12 = 0 so p = -7

Now if second equation have equal roots when b2 -4ac = 0 so p2-4q = 0  put p = -7  then q = 49/4

hence option A

answered by (38 points)

+1 vote