(k+1)x²+2(k+3)x+(k+8)=0
Answers
– A quadratic equation is in the form ax2 +bx +c =0
– To find the nature of roots, first find determinant “D”
– D = b2 – 4ac
– If D > 0, equation has real and distinct roots
– If D < 0, equation has no real roots
– If D = 0, equation has 1 root
(i) 2x2 -3x + 5 =0
Solution:
Here, a= 2, b= -3, c= 5
D = b2 – 4ac
= (-3)2 -4(2)(5)
= 9 – 40
= -31<0
It’s seen that D<0 and hence, the given equation does not have any real roots.
(ii) 2x2 -6x + 3=0
Solution:
Here, a= 2, b= -6, c= 3
D = (-6)2 -4(2)(3)
= 36 – 24
= 12>0
It’s seen that D>0 and hence, the given equation have real and distinct roots.
(iii) (3/5)x2 – (2/3) + 1 = 0
Solution:
Here, a= 3/5, b= -2/3, c= 1
D = (-2/3)2 -4(3/5)(1)
= 4/9 – 12/5
= -88/45<0
It’s seen that D<0 and hence, the given equation does not have any real roots.
(iv) 3x2 – 4√3x + 4 = 0
Solution:
Here, a= 3, b= – 4√3, c= 4
D = (- 4√3)2 -4(3)(4)
= 48 – 48
= 0
It’s seen that D = 0 and hence, the given equation has only 1 real and equal root.
(v) 3x2 – 2√6x + 2 = 0
Solution:
Here, a= 3, b= – 2√6, c= 2
D = (- 2√6)2 – 4(3)(2)
= 24 – 24
= 0
It’s seen that D = 0 and hence, the given equation has only 1 real and equal root.
Answer:
k=1/3
Step-by-step explanation:
(k+1)x²+2(k+3)x+(k+8)=0
a=x+1
b=2(k+3)
c=(k+8)
Δ=b²-4ac=0
Δ=(2(k+3))²-4(k+1)(k+8)=0
=4(k²+9+6k)-4(k²+9k+8)=0
=(k²+9+6k)-(k²+9k+8)=0
=k²+9+6k-k²-9k-8=0
=-3k+1=0
=-3k=-1
=3k=1
=k=1/3
verification
substitute in question
(1/3+1)x²+2(1/3+3)x+(1/3+8)=0
4/3x²+20/3x+25/3=0
4x²+20x+25=0
4x²+10x+10x+25=0
2x(2x+5)+5(2x+5)=0
(2x+5)(2x+5)=0
2x+5=0
x=-5/2