Mathematics, 03.02.2020 21:53, Riohxncho

Problem: for each of the following equations, find the value(s) of the constant _ so that the equation has exactly one solution, and determine
the solution for each value.
(a) _x2 + x + 1 = 0
(b) x2 + _x + 1 = 0
(c) x2 + x + _ = 0
(d) x2 + _x + 4_ + 1 = 0

need .. in advance..

Answers: 3

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Answers

I am going to write k in the place of the constant

The solution is found by making the radicant quantity of the discriminant b^2 - 4(a)(c) = 0, because that is the condition for a quadratic equation to have one and only one real solution.

(a) kx^2 + x + 1 = 0

b^2 - 4(a)(c) = 1^2 -4(k)(1) = 0

1 -4k = 0 => 4k = 1 => k = 1/4

And the solution is [- b +/- √[b^2 - 4ac] ] / 2a = -b / 2a

- 1 / [2 (1/4)] = -2

b) x^2 + kx + 1 = 0

b^2 -4(a)(c) = k^2 -4(1)(1) = k^2 - 4 = 0 => k^2 = 4

k = +/- 2

Solutions: -b/2a

b-1) k = 2 => -2 / 2(1) = - 1
b-2) k = -2 => -(-2) / 2(1) = 1

c) x^2 + x + k =0

b^2 -4(a)(c) = 1 - 4(1)k = 1 - 4k = 0 => 4k = 1 => k =1/4

Solution of the equation: -b / 2a = -1 / 2

d) x^2 +kx +4k + 1

b^2 - 4(a)(c) = k^2 - 4(1)(4k + 1) = k^2 - 16k - 4 = 0

k = 8 + 2√17 and k = 8 -2√17

Solutions

-b/2a = - k / 2a

d-1) - [8 + 2√17] / 2 = -4 - √17

d-2) -[8 - 2√17] / 2 = -4 + √17

answered: Guest

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answered: Guest

step-by-step explanation:

2*5=10

4*15=60

60+10=