The quadratic equation x^2+15|x|+14=0 has | vedclass

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(C) Given the equation $x^2+15|x|+14=0$. Since $x^2 = |x|^2$, we can rewrite the equation as $|x|^2+15|x|+14=0$. Let $|x| = t$, where $t \ge 0$. The e

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no solution

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(C) Given the equation $x^2+15|x|+14=0$. Since $x^2 = |x|^2$, we can rewrite the equation as $|x|^2+15|x|+14=0$. Let $|x| = t$, where $t \ge 0$. The equation becomes $t^2+15t+14=0$. Factoring the quadratic, we get $(t+1)(t+14)=0$. This gives $t = -1$ or $t = -14$. However, we defined $t = |x|$, which must be non-negative $(t \ge 0)$. Since both $-1$ and $-14$ are less than $0$, there is no real value of $x$ that satisfies the equation. Thus, the equation has no solution.

The quadratic equation $x^2+15|x|+14=0$ has

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