Find the value of 'b' if sqrt7 is a root of the equation x^2

Yeah ur posting such wonderful avathars. We go step by step

from the first eq. let p & q be the roots then we know that p+q= - b/a (the eq is of the form ax^2 + bx +c ), so p+q = -a/1 = -a ( eq is x^2+ax-14) and pq = c/a = -14, we are given that one root is sqrt7 then let p=sqrt7 , we know that pq=-14, so (sqrt7* q) = -14 solve and we get q = -2sqrt7, we also know that p =sqrt7 and p+q = -a, substitute the values of p & q and we get a=sqrt7----------------(1)

from the second eq, we r given that it has real and equal roots so it means b^2- 4ac=0 ( for eq of the form ax^2 +bx +c) so as the given eq is of the form x^2 +ax+b=0 , we get a^2 - 4 * 1 * b=0, so a^2=4b, we know that a=sqrt7 ( from -----(1) above) substitute this value of a in a^2=4b and we get b=7/4.

:thanks