Discuss the nature of the roots of the quadrat
ic equation also find its roots if
exist
5x^2-9x-14=0
Answers
Answer:
Step-by-step explanation:
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Answer:
Step-by-step explanation:
Solution: (i) 2x2 -3x + 5 =0
The given quadratic equation is in the form of ax2 +bx +c =0
So a= 2, b= -3, c= 5
We know, determinant (D) = b2 – 4ac
= (-3)2 -4(2)(5)
= 9 – 40
= -31<0
Since D<0, the determinant of the equation is negative, so the expression does not having any real roots.
(ii) 2x2 -6x + 3=0
The given quadratic equation is in the form of ax2 +bx +c =0
So a= 2, b= -6, c= 3
We know, determinant (D) = b2 – 4ac
= (-6)2 -4(2)(3)
= 36 – 24
= 12<0
Since D>0, the determinant of the equation is positive, so the expression does having any real and distinct roots.
(iii) For what value of k (4-k)x2 + (2k+4)x +(8k+1) =0 is a perfect square.
The given equation is (4-k)x2 + (2k+4)x +(8k+1) =0
Here, a= 4-k, b= 2k+4, c= 8k+1
The discriminate (D) = b2 – 4ac
= (2k+4)2 – 4(4-k)(8k+1)
= (4k2+16+ 16k) -4(32k+4-8k2-k)
= 4(k2 +8k2+4k-31k+4-4)
=4(9k2-27k)
D = 4(9k2-27k)
The given equation is a perfect square
D= 0
4(9k2-27k) = 0
9k2-27k=0
Taking out common of of 3 from both sides and cross multiplying
= k2 -3k =0
= K (k-3) =0
Either k=0
Or k =3
The value of k is to be 0 or 3 in order to be a perfect square.
(iv) Find the least positive value of k for which the equation x2+kx+4=0 has real roots.
The given equation is x2+kx+4=0 has real roots
Here, a= 1, b= k, c= 4
The discriminate (D) = b2 – 4ac ≥0
= k2 – 16 ≥ 0
= k≤ 4 ,k≥-4
The least positive value of k =4 for the given equation to have real roots.