Question

kx^2 - 6x + 4 has real and equal roots find the value of k

Answer 1

Discriminant = ${b}^{2}$ - 4ac

Given,
Roots are real and equal

$\therefore$ Discriminant => ${b}^{2}$ - 4ac = 0

a = k
b = -6
c =4

$= >{b}^{2} - 4ac = 0 \\ \\ = > 36 - 4 \times k \times 4 = 0 \\ \\ = > 36 - 16k = 0 \\ \\ = > - 16k = - 36 \\ \\ = > k = \frac{36}{16} \\ \\ \therefore \: k = \frac{9}{4}$



$<b>Verification</b>$

Discriment = b^2 - 4ac
= 36 - 4 × $\frac{9}{4}$ × 4
=36 - 36
=0

$\since$ Discriminant = 0

therefore the roots will be real and equal

Answer 2

Answer:

It is given that kx^2 -6x+4 has real and equal roots.

Therefore,D=0

b^2-4ac=0

Here,a=K, b=-6 and c=4

Putting the values,we have

(-6)^2- 4(K)(4)=0

36-16K=0

36=16K

36/16=K

9/4=K

Hence,the value of K is 9/4