Find Values of x for y=2x²-50 Where y < 0: A Quadratic Exploration

To solve the inequality $2x^2 - 50 < 0$, we follow these steps:

• Set the quadratic equation equal to zero to find the roots: $2x^2 - 50 = 0$.

• Rearrange and solve for $x$:

$\begin{aligned} 2x^2 - 50 &= 0\\ 2x^2 &= 50\\ x^2 &= 25\\ x &= \pm \sqrt{25}\\ x &= \pm 5 \end{aligned}$

These roots, $x = -5$ and $x = 5$, are where the function $y = 2x^2 - 50$ is equal to zero.

We now examine the intervals determined by these roots to find where the function is negative:

• $x < -5$

• $-5 < x < 5$

• $x > 5$

Since the quadratic is an upward opening parabola (coefficient of $x^2$ is positive), it attains its minimum value between its roots and increases outside them.

Testing a point in each interval:

• (For $x = 0$ in the interval $-5 < x < 5$): $y = 2(0)^2 - 50 = -50 < 0$.

• (Other intervals will be positive) such as $x = -6$ or $x = 6$, will have $y > 0$.

Thus, the function is negative in the interval $-5 < x < 5$.

Therefore, the values of $x$ that satisfy $2x^2 - 50 < 0$ are:

$-5 < x < 5$.