Question

1.Find the values of "k" for which the following equations has one roots
kx²+kx-2=0

2.Find the value of "a" for which the following equations has no roots
x²-ax+a+3=0

3.Equation 4x²+a=2x has real roots and Discriminants value is lower than 1

Please help me

Answer 1

Answer:

1) k = -8

2) -2 < a < 6

3) a = 1/4

Step-by-step explanation:

To have equal roots

Discriminant = D = 0

D = b² - 4ac

ax² + bx + c = 0

1)

kx²+kx-2=0

a = k  b = k c = -2

k² - 4k(-2) = 0

k² + 8k = 0

k + 8 = 0    ( as k can not be 0 )

k = -8

2)

x²-ax+a+3=0

x² - ax  +(a+3) = 0

to have no roots

D < 0

(-a)² - 4(1)(a+3) < 0

=> a² - 4a - 12 < 0

=> a² - 6a + 2a - 12 < 0

=> a(a-6) + 2(a -6) < 0

=> (a+2)(a-6) < 0

this is possible if one vlaue is -ve & other is + v2

Case 1  a + 2> 0  & a - 6<0

=> a > -  2  & a < 6

-2 < a < 6

case 2

a + 2 < 0   & a - 6 > 0

=> a < -2 & a > 6

not possible

so -2 < a < 6

3)

4x²+a=2x

4x² -2x + a = 0

(-2)² - 4(4)(a) = 0

=> 4 - 16a = 0

=> a = 4/16

=> a = 1/4

D = 0 so its lower than 1