Question

if the roots of 2x^2 +kx+3= are real and equal then k=​

Answer 1

Step-by-step explanation:

2x² +kx+3=0

If roots are real and equal then discriminant value of equation, b²-4ac=0

b=k, a=2, c=3

b²-4ac=0

k²-4(2)(3)=24

k=√24

k=2√6.

Answer 2

★︎︎︎Aɴsᴡᴇʀ :-

• k = 2√6

★︎︎︎Qᴜᴇsᴛɪᴏɴ :-

• ɪf the roots of 2x² +kx+3= are real and equal then k= ?

★︎︎︎Gɪᴠᴇɴ :-

• Roots are real and equal.
• 2x² + kx +3

here...... Let coefficient of

• x² be 'a'
• x be 'b'

constant be 'c'

★︎︎︎Cᴏɴᴄᴇᴘᴛ :-

• The sum of zeroes of quadratic polynomial is -b/a
• The product of zeroes of quadratic polynomial is c/a

★︎︎︎Sᴏʟᴜᴛɪᴏɴ :-

Let the two zeroes be α and β

So, α+β = -b/a

α+β = -k/2

➜ Since , α=β

• 2β = -k/2 -----------------(1)

Again, β.α = c/a

• So, β² = 3/2 ----------------------(2)

From equation (1) , β = -k/4

↪putting this value of β in equation (2).....

$({ \frac{ - k}{4}) }^{2} = \frac{3}{2} \\ \\ \frac{ {k}^{2} }{16} = \frac{3}{2} \\ \\ {k}^{2} = 24 \\ \\ k = \sqrt{2 \times 2 \times 2 \times 3} \\ \\ k = 2 \sqrt{6}$

★Vᴇʀɪғɪᴄᴀᴛɪᴏɴ :-

2x² + 2√6x + 3 = 0

$roots \: = \frac{ - b \binom{ + }{ - } \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ = \frac{ - 2 \sqrt{6} \binom{ + }{ - } \sqrt{( {2 \sqrt{6} )}^{2} - 4(2)(3) } }{2(2)} \\ \\ = \frac{ - 2 \sqrt{6} \binom{ + }{ - } \sqrt{24 - 24} }{4} \\ \\ roots \: = \frac{ - 2 \sqrt{6} + 0 }{4} ..... \frac{ - 2 \sqrt{6} - 0}{4} \\ \\ roots \: = \frac{ - \sqrt{6} }{2} ...... \frac{ - \sqrt{6} }{2}$

Sᴏ, ᴛʜᴇ ʀᴏᴏᴛs ᴀʀᴇ ᴇǫᴜᴀʟ.....

ᴡᴇ ᴄᴀɴ ɴᴏᴡ ᴄᴏɴᴄʟᴜᴅᴇ ᴛʜᴀᴛ

k = 2√6 (verified)

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