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Find the value of K for which the quadratic equation x^-4x+k=0has distinct real roots

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Answered by prabhat00127
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Answered by Anonymous
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Question:

Find the value of k for which the quadratic equation x² - 4x + k = 0 has real and distinct roots.

Answer:

k € (-∞,4)

Note:

• An equation of degree 2 is know as quadratic equation .

• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.

• The maximum number of roots of an equation will be equal to its degree.

• A quadratic equation has atmost two roots.

• The general form of a quadratic equation is given as , ax² + bx + c = 0 .

• A quadratic equation has atmost two roots .

• The discriminant of the quadratic equation is given as , D = b² - 4ac .

• If D = 0 , then the quadratic equation would have real and equal roots .

• If D > 0 , then the quadratic equation would have real and distinct roots .

• If D < 0 , then the quadratic equation would have imaginary roots .

Solution:

The given quadratic equation is ;

x² - 4x + k = 0

Clearly , we have ;

a = 1

b = -4

c = k

We know that ,

The quadratic equation will have real and distinct roots if its discriminant is greater than zero .

=> D > 0

=> (-4)² - 4•1•k > 0

=> 16 - 4k > 0

=> 16 > 4k

=> 4k < 16

=> k < 16/4

=> k < 4

=> k € (-∞,4)

Hence,

The required values of k are (-∞,4) .

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