Question

find the value of n (n+3) x^2-(5-n) x+1=0 has coincident roots​

Answer 1

Question :

Find the value of n for which the quadratic equation (n+3)x² - (5-n)x + 1 = 0 has coincident roots .

Answer :

n = 1 , 13

Note:

★ The possible values of the variable which satisfy the equation are called its roots or solutions .

★ A quadratic equation can have atmost two roots .

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

★ The discriminant , D of the quadratic equation ax² + bx + c = 0 is given by ;

D = b² - 4ac

★ If D = 0 , then the roots are real and equal .

★ If D > 0 , then the roots are real and distinct .

★ If D < 0 , then the roots are unreal (imaginary) .

Solution :

Here ,

The given quadratic equation is ;

(n+3)x² - (5-n)x + 1 = 0

Now ,

Comparing the given quadratic equation with the general quadratic equation ax² + bx + c = 0 , we have b;

a = n + 3

b = -(5-n) = n - 5

c = 1

For equal or coincident roots , the discriminant of the given quadratic equation must be zero .

Thus ,

=> D = 0

=> b² - 4ac = 0

=> (n - 5)² - 4•(n + 3)•1 = 0

=> n² - 10n + 25 - 4n - 12 = 0

=> n² - 14n + 13 = 0

=> n² - n - 13n + 13 = 0

=> n(n - 1) - 13(n - 1) = 0

=> (n - 1)(n - 13) = 0

=> n = 1 , 13

Hence , n = 1 or 13 .

Answer 2

Answer:

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