Question

find the value of k, for the following quadratic equation, so that they have two equal roots 2x+kx+6=0​

Answer 1

EXPLANATION :

As we know the QUADRATIC EQUATION FORMULA

$\sf \: x = \frac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a}$

we can see

$\sf \sqrt{ {b}^{2} - 4ac } \: \: should \: be \: zero \: to \: get \: equal \: roots$

$\sf \sqrt{ {b}^{2} - 4ac } = 0$

$\implies \sf {b}^{2} = 4ac$

putting value

$\implies \sf {k}^{2} = 4 \times 6 \times 2$

${ \boxed{\sf \: Hance \: k \: should \: be \: equal \: to \: \sqrt{48} \: to \: get \: equal \: roots}}$

Answer 2

Step-by-step explanation:

For equal roots

D = 0

b² - 4ac = 0

k² - 4*2*6 = 0

k² - 48 = 0

k² = 48

k = √48

k = 4√3

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