Quadratic Equation - 4 · KnowledgeBin.org

Consequently the original equation is equivalent to the collection of equations,

We find that

Q-5: If aÎz and the equation (x-a) (x-10) +1=0 has integral roots, then what is the value of ‘a’?

Solution:

As both x and a are integers and hence given equation implies that either x-a =1 and x-10= -1 or x-a = -1 and x-10=1.

x=9 and hence a=8 or x=11 and a=12.

Therefore possible values of a’s are 8, 12.

Q-6: Find the values of ‘a’ for which inequality is true for at least one

Solution: The required condition will be satisfied if – The quadratic expression (quadratic in tanx).
1)      f (x) = has positive discriminant and ,
2)      At least one root of f(x) =0 is positive as tanx>0, for all
For (1) discriminant >0

For (2) we first find the condition, that both the roots of
(t=tanx) are non positive for which,
Sum of roots<0, product of roots 0
- (a+1) <0 and – (a-3) 0 -1 <a 3
Condition (2) will be fulfilled of a -1or a>3            - - - - - - - - - -(b)
Required value of a is given by intersection of (a) and (b)

Hence

Q-7: Solve the equation- where (x) and [x] are the integer just less than or equal to x and just greater than or equal to x respectively.

Case 1: If x I,

Then (x) = [x] - - - - - - - - (1),

Case 2: if x l
Then (x) = [x]-1

Hence the solution of original equation is, x=0, .

Q-8: If the equation has real roots a, b, c being real numbers and if m and n are real number such that m2>n>0 then prove that the equation has real roots.

Solution: Since roots of the equation are real

and discriminant of

Hence roots of equation are real.

Q-9 If graph of function is strictly above the x-axis then show that -15<a<-2
Solution: Y has to be positive
16x2+8(a+5) x-(7a+5) = +ve
Since sign of first term is +ve therefore the expression will be +ve if D<0.

Q-10: For real roots what is the solution of the equation
Solution:
Divide by where t is +ve being exponential function.

The other value is rejected as t is +ve.