Question

Q+. Determine the nature of roots of the
following Quadratic equation
1) - 2x²-4x+3=0
2) x ² + lox+39=0
3) k^2x^2 + kx+1=0
4)2x² + x -1 = 0​

Answer 1

Step-by-step explanation:

We've to use the formula in each part:-

D=(b)^2 -4ac

(1) D= b^ -4ac

$= { - 4}^{2} - 4 \times ( - 2) \times 3$

$= 16 + 24$

$= 40$

'cause D>0

Therefore, The roots're real an' unequal

(2) D=b^2 - 4ac

= 10^2 -4*1*39

= 100-156

= -56

'cause D<0

Therefore

the roots are not real

(3(K^2)(x^2) +kx +1 =0

D = k^2 -4* k^2* 1

= k^2 -4(k^2)

= -3ksq.

'cause D<0

therefore,

the roots are not real.

(4)

D= 1^2 -4*(2)*(-1)

= 1+8

= 9

'cause D>0

therefore,

The roots are real and unequal.

Answer 2

solution:-

Rule:-

if ax²+bx+c = 0 be a quadratic equation then

b²>4ac => root are real unequal

b²= 4ac => root are real equal

b² < 4ac => root are imaginary

(1) b²=16 ,4ac= -12 => real distinct ( unequal)

(2) b²= 100 ,4ac= 156 => imaginary

(3) b²= k², 4ac = 4k²

here if k<0 => k² is positive

k>0 => k² also positive=> k² < 4k²

hence root are imaginary

note:- if a,b,are c are not given in their numerical form then the may be negative or may be positive. assuming this before calculation.

(4) b²= 1 , 4ac = - 8 => real distinct