Question

$\red∙ \purple∘\green □ \sf \: Question \: of \: the \: day \: \green□\purple∘ \red\bullet$
Write the values of $k$ for which $2x²+kx-8=0$ has real roots. ​

Answer 1

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✤ Required Answer:

✒ GiveN:

• Polynomial given = 2x² + kx - 8 = 0
• The polynomial have real roots.

✒ To FinD:

• Value of k for satisfying above condition.

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✤ How to solve?

Firstly, We can determine whether the roots of polynomial real or imaginary or real and equal by judging the Discriminant(D). It can be found by:

➝ D = b² - 4ac

And, Here are the condition:

• D > 0 ⇛ Roots are real.
• D = 0 ⇛ Roots are real and equal.
• D < 0 ⇛ Roots are imaginary.

☃️ So, Let's solve the equation.....

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✤ Solution:

We have,

• Polynomial P(x) = 2x² + kx - 8 = 0

Compare with ax² + bx + c

Finding Discriminant,

➝ D = b² - 4ac

➝ D = k² - 4 × 2 × (-8)

➝ D = k² + 16

Now,

According to question, The roots are real. Here, There are two cases:

• D > 0
• D = 0

So, We will take both the cases as D ≥ 0

Then,

➝ D ≥ 0

➝ k² + 16 ≥ 0

➝ k² ≥ - 16

➝ k ≥ √-16

➝ k ≥ √16 × √-1

➝ k ≥ 4i [ i refers to iota which is = √-1 ]

So, The Required answer is:

$\therefore {\boxed{ \sf{k \geqslant 4 \iota}}}$

All values of K greater than or equal to 4i

☀️ Hence, solved !!

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Answer 2

Step-by-step explanation:

Polynomial P(x) = 2x² + kx - 8 = 0

Compare with ax² + bx + c

Finding Discriminant,

➝ D = b² - 4ac

➝ D = k² - 4 × 2 × (-8)

➝ D = k² + 16

Now,

According to question, The roots are real. Here, There are two cases:

D > 0

D = 0

So, We will take both the cases as D ≥ 0

Then,

➝ D ≥ 0

➝ k² + 16 ≥ 0

➝ k² ≥ - 16

➝ k ≥ √-16

➝ k ≥ √16 × √-1

➝ k ≥ 4i [ i refers to iota which is = √-1 ]

So, The Required answer is:

$k \geqslant 4t$

All values of K greater than or equal to 4i

☀️ Hence, solved !!