Question

Find the least value of k fir which the quadratic equation -3x^2-6x+k=0 has real roots

Answer 1

Step-by-step explanation:

if the equation has equal real roots then ,D=0

also, b^2-4ac=0

here b=-6, a=-3 , c=k

(-6)^2-4×-3×k =0

36+12k=0

12k=-36

k=-36÷12

k=-3

Answer 2

Question:

Find the value of k for which the quadratic equation -3x² - 6x + k = 0 has equal roots.

Answer:

k = -3

Note:

• An equation of degree 2 is know as quadratic equation .

• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.

• The maximum number of roots of an equation will be equal to its degree.

• A quadratic equation has atmost two roots.

• The general form of a quadratic equation is given as , ax² + bx + c = 0 .

• The discriminant of the quadratic equation is given as , D = b² - 4ac .

• If D = 0 , then the quadratic equation would have real and equal roots .

• If D > 0 , then the quadratic equation would have real and distinct roots .

• If D < 0 , then the quadratic equation would have imaginary roots .

Solution:

The given quadratic equation is ;

-3x² - 6x + k = 0

Clearly , we have ;

a = -3

b = -6

c = k

We know that ,

The quadratic equation will have equal roots if its discriminant is equal to zero .

=> D = 0

=> (-6)² - 4•(-3)•k = 0

=> 36 + 12k = 0

=> 12k = -36

=> k = -36/12

=> k = -3

Hence,

The required values of k is - 3.