Consider the equation `|x + 2| + |x - 2| = 4 - px` The equation has one solution if p is

Consider the equation |x^2 - 4 |x| +3|=p (A) for p = 2 the equation has four solutions(B) for p = 2 the equation has eight solutions(c) there exists valueonly one real of p for which the equation has odd number of solutions(D) sum of roots of the equation is zero irrespective of value of p

Consider the equation |x^(2)-2x-3|=m , m in R . If the given equation has two solutions, then:

Consider the equation |x^(2)-2x-3|=m , m in R . If the given equation has two solutions, then:

Consider the equation |x-1|-1|=px+c. Match the number of solution with the value of p and c.

Find the value(s) of p for the following pair of equations: -x+ py =1 and px- y=1 , if the pair of equations has no solution.

If -3 is a root of the equation x^2+ px + 6 = 0 and the equation x^2 +px +q = 0 has equal roots, find p,q .

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

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

If one root of the equation x^(2) + px + 12 = 0 is 4, while the equation x^(2)+ px + q = 0 has equal roots, then the value of 'q' is