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x^(2)+kx-14=0
In the given equation,
k is a constant. The equation has solutions at 7 and -2 . What is the value of
k ?
Choose 1 answer:
(A) -9
(B) -5
(C) 5
(D) 9

$x^{2}+kx-14=0$ $\newline$ In the given equation, $k$ is a constant. The equation has solutions at $7$ and $-2$ . What is the value of $k$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-9$ $\newline$ (B) $-5$ $\newline$ (C) $5$ $\newline$ (D) $9$

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Solve trigonometric equations

Full solution

Q.

x^{2}+kx-14=0

\newline

In the given equation,

k

is a constant. The equation has solutions at

7

and

-2

. What is the value of

k

?

\newline

Choose

1

answer:

\newline

(A)

-9

\newline

(B)

-5

\newline

(C)

5

\newline

(D)

9

Relationships between coefficients: Since the solutions to the quadratic equation are given as $7$ and $-2$ , we can use the fact that the solutions to a quadratic equation of the form $ax^2 + bx + c = 0$ are related to the coefficients by the relationships sum of roots = $-b/a$ and product of roots = $c/a$ . In this case, $a = 1$ , so the sum of the roots is $-k/1$ and the product of the roots is $-14/1$ .
Calculate sum of roots: The sum of the roots is $7 + (-2) = 5$ . According to the relationship sum of roots $= -b/a$ , we have $5 = -k/1$ . Therefore, $k = -5$ .
Calculate product of roots: The product of the roots is $7 \times (-2) = -14$ . According to the relationship product of roots $= \frac{c}{a}$ , we have $-14 = \frac{-14}{1}$ . This confirms that the product of the roots is consistent with the constant term of the quadratic equation.
Determine value of $k$ : Since the sum of the roots is $5$ and we have found that $k = -5$ , we can conclude that the value of $k$ is indeed $-5$ . This corresponds to option (B) in the given choices.

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