Find all values of $k$ for which the quadratic equation has two real solutions.
$2 x^{2} + 7 x + k = 0$
Write your answer as an equality or inequality in terms of $k$ .

Answer

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Answer

$k < \frac{49}{8}$ is the final answer. The quadratic equation has two real solutions for all values of $k$ that are less than $\frac{49}{8}$ .

Steps

Step 1 ：Given the quadratic equation $2 x^{2} + 7 x + k = 0$ , we need to find all values of $k$ for which the equation has two real solutions.

Step 2 ：A quadratic equation has two real solutions if and only if the discriminant of the equation is greater than 0. The discriminant of a quadratic equation in the form $a x^{2} + b x + c = 0$ is given by $b^{2} - 4 a c$ .

Step 3 ：In this case, $a = 2$ , $b = 7$ , and $c = k$ . So, we need to find all values of $k$ such that $7^{2} - 4 * 2 * k > 0$ .

Step 4 ：Solving the inequality $49 - 8 k > 0$ gives us $k < \frac{49}{8}$ .

Step 5 ： $k < \frac{49}{8}$ is the final answer. The quadratic equation has two real solutions for all values of $k$ that are less than $\frac{49}{8}$ .

Find all values of k for which the quadratic equation has two real solutions.2x2+7x+k=0Write your answer as an equality or inequality in terms of k.

Answer

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Find all values of $k$ for which the quadratic equation has two real solutions.
$2 x^{2} + 7 x + k = 0$
Write your answer as an equality or inequality in terms of $k$ .