Question

Hello

Question

For what vaule of k , tge roots of the equation
${x}^{2} \: + 4x + k \: = 0 \: are \: real \:$

$\: ans \: = \: 4$
I am not sure

Explain it


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Answer 1

For a quadratic equation, ax² + bx + c

The roots are given by,

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

Since a, b are real. The only term that can decide is the Discriminant ( b² - 4ac)

If it's less than 0, Negative number under root will be imaginary. Hence, For real roots we need b² - 4ac to be greater than or even equal to 0.

Given equation, x² + 4x + k = 0

here a = 1, b = 4, c = k

For roots to be real,

${b}^{2} - 4ac \: \geqslant 0 \\ \\ {4}^{2} - 4(1)(k) \geqslant = 0 \\ \\ 4(4 - k) \geqslant 0 \\ \\ 4 - k \geqslant 0 \\ \\ 4 \geqslant k$

Therefore, k can assume any value less than or equal to 4.

Further if your question asks for equal & real roots, k = 4 gives you real & equal roots.

Answer 2

Answer:

ello ❤here is your answer!

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plz refer attachment✌