Question

Find the values of k for which the quadratic equation 9x 2+3kx+4 has real roots

Answer 1

from formula
roots of quadratic equation are=
$\frac{ - 3k + - \sqrt{ {(3k)}^{2} - 4 \times 9 \times 4} }{2a}$
if roots are real then k²-16>0
or (k-4)(k+4)>0
then
case I case II
k-4>0 & k+4>0. k-4<0 & k+4<0
or k>4. or k>-4. or k<4. or k<-4
then values varies from k>-4 or k<4.

Answer 2

Question:

Find the value of k for which the quadratic equation 9x² - 3kx + 4 = 0 has real roots.

Answer:

k € [-∞,-4]U[4,∞]

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 .

• A quadratic equation has atmost two roots .

• 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 ;

9x² - 3kx + 4 = 0

Clearly , we have ;

a = 9

b = -3k

c = 4

We know that ,

The quadratic equation will have real roots if its discriminant is greater than or equal to zero .

=> D ≥ 0

=> (-3k)² - 4•9•4 ≥ 0

=> 9k² - 9•16 ≥ 0

=> 9•(k² - 16) ≥ 0

=> k² - 16 ≥ 0

=> (k+4)(k-4) ≥ 0

=> k € [-∞,-4]U[4,∞]

Hence,

Hence,The required values of k are [-∞,-4]U[4,∞] .