find value of k having real and equal root : x²+(k-3)x+k=0
Answers
Step-by-step explanation:
Compare the equation x^2+(k-3)x+k=0 with the quadratic equation ax^2+bx+c=0.we get,
a= 1, b=(k-3) and c = k.
Discriminant =0
b^2-4ac=0
=>(k-3)^2-4*1*k=0
=>k^2-6k+9-4k=0
=>k^2-10k+9=0
=>k^2-(9+1)k+9=0
=>k^2-9k-k+9=0
=>k(k-9)-1(k-9)=0
=>(k-9)(k-1)=0
either, k-9=0, k=9
or, k-1=0, k= 1
The value of k is 9 and 1
Given : x² + (k-3)x + k = 0
To Find : value of k having real and equal root
Solution :
Quadratic equation is of the form ax²+bx+c=0 where a , b and c are real also a≠0.
D = b²-4ac is called discriminant.
D >0 roots are real and distinct
D =0 roots are real and equal
D < 0 roots are imaginary ( not real ) and different
x² + (k-3)x + k = 0
a = 1 , b = k - 3 , c = k
(k - 3)² - 4(1)(k) = 0
=> k² -6k + 9 - 4k = 0
=> k² - 10k + 9 = 0
Using middle term split
=> k² - 9k - k + 9 = 0
=> k(k -9) - 1(k - 9) = 0
=> (k - 9)(k - 1) = 0
=> k = 9 , k = 1
Values of k are 1 and 9
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