Question

For what value of k the kx^2-5x+k=0 has real and equal roots

Answer 1

1- b^2 - 4ac = (-5)^2 - 4(3)(2) = 25 - 24 = 1

There are two unequal rational roots.

2- b^2 - 4ac = (-40)^2 - 4(2)(25) = 1600 - 200 = 1400

Where is the k?

3- b^2 - 4ac = (-4)^2 - 4(2)(3) = 16 - 24 = -8

The roots are complex conjugate, not equal roots.

4- sum = -b/a = -(-5)/2 = 5/2

product = c/a = 4/2 = 2

5- Let n and n + 1 be consecutive integers.

n^2 + (n + 1)^2 = 25

n^2 + n^2 + 2n + 1 = 25

2n^2 + 2n - 24 = 0

n^2 + n - 12 = 0

(n + 4)(n - 3) = 0

n = -4 or n = 3

n + 1 = -3 or n + 1 = 4

Since the integers are positive, the integers are 3 and 4

I hope this will help you

If not then comment me

Answer 2

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