Question

if 2 isa root of the equation x^2+bx+c=0,and the equation x2+bx+q=0,has equal roots, then q=?? Its urgent. pls do fast

Answer 1

Putting p=2 in first equation-->
(2)^2+b×(2)+c=0
4+2b+c=0
2b+c= -4 {equation 1}
Now, putting p=2 in second equation-->
(2)×2+b(2)+q=0
4+2b+q=0
2b+q= -4. {equation 2}
Now eq. 1 = -4 and eq. 2= -4
Thus eq. 1 = eq. 2
--> 2b+c=2b+q
2b gets cancelled
Thus
c=q or q=c is your answer
hope this helps, plz mark as brainliest

Answer 2

SOLUTION :  

Option (c) is correct : 16

Given :  2 is the root of x² + bx + 12 = 0 ………..(1)

and x² + bx + q = 0 has equal roots………(2)

Since, x = 2  is a root of equation (1) so it will satisfy the equation.

On putting x = 2  in equation (1)

x² + bx + 12 = 0  

(2)² + 2b + 12 = 0

4 + 2b + 12 = 0

16 + 2b = 0

2b = - 16

b = -16/2  

b = - 8

On putting b = - 8  in equation (2)

x² + bx + q = 0

x² + (- 8)x + q = 0

x² - 8x + q = 0

On comparing the given equation with ax² + bx + c = 0  

Here, a = 1 , b = - 8 , c =  

D(discriminant) = b² – 4ac

D = (- 8)² - 4(1)(q)

D = 64 - 4q

D = 0 ( Equal  roots given)

64 - 4q = 0

64 = 4q

q = 64/4  

q = 16

Hence the value of q is 16 .

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