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In the given equation, $k$ is a constant. For what value of $k$ does the equation have exactly one distinct real solution? $\newline$ Choose $1$ answer: $\newline$ (A) $-\frac{25}{8}$ $\newline$ (B) $-\frac{5}{4}$ $\newline$ (C) $\frac{5}{4}$ $\newline$ (D) $\frac{25}{8}$

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Operations with rational exponents

Full solution

Q. In the given equation,

k

is a constant. For what value of

k

does the equation have exactly one distinct real solution?

\newline

Choose

1

answer:

\newline

(A)

-\frac{25}{8}

\newline

(B)

-\frac{5}{4}

\newline

(C)

\frac{5}{4}

\newline

(D)

\frac{25}{8}

Quadratic Equation Real Solution: A quadratic equation has exactly one distinct real solution when its discriminant is zero. The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by $b^2 - 4ac$ .
Given Equation and Parameters: For the given equation $2x^2 + 5x - k = 0$ , $a = 2$ , $b = 5$ , and $c = -k$ . We will set the discriminant equal to zero and solve for $k$ .
Calculate Discriminant: The discriminant is $b^2 - 4ac$ , so we have $5^2 - 4(2)(-k) = 0$ .
Discriminant Calculation: Calculate the discriminant: $5^2 - 4(2)(-k) = 25 + 8k = 0$ .
Solve for k: Solve for $k$ : $8k = -25$ .
Final Solution for $k$ : Divide both sides by $8$ to get $k$ : $k = -\frac{25}{8}$ .

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f(x)=\frac{6 x+5}{2-\sqrt{14+5 x}}

\newline

We want to find

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\newline

What happens when we use direct substitution?

\newline

1

\newline

(A) The limit exists, and we found it!

\newline

(B) The limit doesn't exist (probably an asymptote).

\newline

(C) The result is indeterminate.

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