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kx^(2)-3x=-2
In the given equation,
k is a constant. For which values of
k will the equation have no real solutions?
Choose 1 answer:
(A)
k < -(9)/(8)
(B)
k > -(9)/(8)
(C)
k < (9)/(8)
(D)
k > (9)/(8)

$k x^{2}-3 x=-2$ $\newline$ In the given equation, $k$ is a constant. For which values of $k$ will the equation have no real solutions? $\newline$ Choose $1$ answer: $\newline$ (A) k<-\frac{9}{8} $\newline$ (B) k>-\frac{9}{8} $\newline$ (C) k<\frac{9}{8} $\newline$ (D) k>\frac{9}{8}

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Is (x, y) a solution to the system of linear inequalities?

Full solution

Q.

k x^{2}-3 x=-2

\newline

In the given equation,

k

is a constant. For which values of

k

will the equation have no real solutions?

\newline

Choose

1

answer:

\newline

(A)

k<-\frac{9}{8}

\newline

(B)

k>-\frac{9}{8}

\newline

(C)

k<\frac{9}{8}

\newline

(D)

k>\frac{9}{8}

Rewrite in Standard Form: First, we need to rewrite the equation in standard quadratic form, which is $ax^2 + bx + c = 0$ . We do this by adding $2$ to both sides of the equation. $\newline$ $kx^2 - 3x + 2 = 0$
Use Discriminant: Next, we determine the conditions for which a quadratic equation has no real solutions by using the discriminant. The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by $b^2 - 4ac$ . If the discriminant is less than zero, the equation has no real solutions. $\newline$ For our equation, $a = k$ , $b = -3$ , and $c = 2$ . So the discriminant is $(-3)^2 - 4(k)(2)$ .
Calculate Discriminant: Now, we calculate the discriminant: $\newline$ Discriminant = $(-3)^2 - 4(k)(2) = 9 - 8k$ .
Set Up Inequality: For the equation to have no real solutions, the discriminant must be less than zero: 9 - 8k < 0.
Solve for $k$ : We solve the inequality for $k$ :-8k < -9k > \frac{9}{8}.
Final Answer: The inequality k > \frac{9}{8} means that for any value of $k$ greater than $\frac{9}{8}$ , the equation will have no real solutions. Therefore, the correct answer is: $\newline$ (D) k > \frac{9}{8}.

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