1.

3. Given that Lcm (9236) = 185 then find4CE (9 1226 ) ter dhff A-​

Answer»

Answer:

This IMPLIES that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just ONE SOLUTION instead of the usual two distinct solutions, then the two solutions must be same i.e. DISCRIMINANT = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.


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