Question

find k for which the following equation has real & distinct roots kx (x-2) +6 =0​

Answer 1

Answer:

Given,

kx(x−2)+6=0

kx

2

−2kx+6=0

Since the roots are equal,

⇒b

2

=4ac

(−2k)

2

=4(k)(6)

4k

2

=4k(6)

∴k=6

Step-by-step explanation:

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Answer 2

Answer:-

• Simplify first

$\\ \sf\longmapsto kx(x-2)+6=0$

$\\ \sf\longmapsto kx^2-2kx+6=0$

• a=k
• b=2k
• c=6

If it has real and distinct roots then

$\\ \sf\longmapsto D=0$

$\\ \sf\longmapsto b^2-4ac=0$

$\\ \sf\longmapsto (2k)^2-4(k)(6)=0$

$\\ \sf\longmapsto 4k^2-24k=0$

$\\ \sf\longmapsto 4k^2=24k$

$\\ \sf\longmapsto 4k=24$

$\\ \sf\longmapsto k=\dfrac{24}{4}$

$\\ \sf\longmapsto k=6$