For what value of k, does the following quadratic equation have equal roots?
(i) k²x
²-2(2k-1)x+4=0
(ii) (k+4)x²+(k+1)x+1=0
Answers
Answer:
zhzjz. znausjsbsznKIaosbs znzmsks
Step-by-step explanation:
Given :-
(i) k²x²-2(2k-1)x+4=0
(ii) (k+4)x²+(k+1)x+1=0
To find :-
For what value of k, does the following quadratic equation have equal roots?
Solution :-
(i)
Given quadratic equation is k²x²-2(2k-1)x+4=0
On comparing this with the standard quadratic equation ax²+bx+c = 0
a = k²
b = -2(2k-1)
c = 4
Given that
Given quadratic equation has equal roots then
The discriminant of the given equation is zero
We know that
The discriminant of ax²+bx+c = 0 is D = b²-4ac
So we have
b²-4ac = 0
=> [-2(2k-1)]²-4(k²)(4) = 0
=> 4(2k-1)²-16k² = 0
=> 4[(2k)²-2(2k)(1)+1²] -16k² = 0
=> 4(4k²-4k+1) -16k² = 0
=> 16k²-16k+4-16k² = 0
=> (16k²-16k²)-16k+4 = 0
=> 0-16k+4 = 0
=> -16k +4 = 0
=> -16k = -4
=> 16k = 4
=> k = 4/16
=> k = 1/4
Therefore, k = 1/4
-----------------------------------------------------
ii)
Given quadratic equation is
(k+4)x²+(k+1)x+1=0
On comparing this with the standard quadratic equation ax²+bx+c = 0
a = k+4
b = k+1
c = 1
Given that
Given quadratic equation has equal roots then
The discriminant of the given equation is zero
We know that
The discriminant of ax²+bx+c = 0 is D = b²-4ac
So we have
b²-4ac = 0
=> (k+1)²-4(k+4)(1) = 0
=> k²+2k+1-4(k+4) = 0
=> k²+2k+1-4k-4 = 0
=> k²-2k-3 = 0
=> k²+k-3k -3 = 0
=> k(k+1)-3(k+1) = 0
=> (k+1)(k-3) =0
=> k+1 = 0 or k-3 = 0
=> k = -1 or k = 3
Therefore, k = -1 and 3
Answer:-
(i) The value of k for the given problem is 1/4
(ii) The values of k for the given problem are -1 and 3
Used formulae:-
- The standard quadratic equation is ax²+bx+c = 0.
- The discriminant of ax²+bx+c = 0 is D= b²-4ac
- If an equation has equal roots then the discriminant is equal to zero.
Points to know :-
- The discriminant of ax²+bx+c = 0 is D= b²-4ac
- If D > 0 ,then it has two distinct and real roots
- If D < 0 ,then it has no real roots
- If D= 0 ,then it has equal and real roots.