Given: $$ (x+5)^{2}=1-p^{2} $$ Calculate the values of $$ p $$ for which the roots of the equation are non-rea

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To determine the values of $$ p $$ for which the roots of the equation $$ (x+5)^{2}=1-p^{2} $$ are non-real, we need to analyze the right side of the equation. 1. **Rearranging the equation**: We can rewrite the equation as: $$ (x+5)^{2} - (1 - p^{2}) = 0 $$ This is a quadratic equation in the form $$ ax^2 + bx + c = 0 $$. 2. **Condition for non-real roots**: The roots of a quadratic equation are non-real if the discriminant is less than zero. The discriminant $$ D $$ for the equation $$ (x+5)^{2} - (1 - p^{2}) = 0 $$ can be expressed as: $$ D = b^{2} - 4ac $$ Here, $$ a = 1 $$, $$ b = 0 $$, and $$ c = -(1 - p^{2}) $$. Therefore, the discriminant becomes: $$ D = 0^{2} - 4 \cdot 1 \cdot (-(1 - p^{2})) = 4(1 - p^{2}) $$ 3. **Setting the condition for non-real roots**: We need to find when $$ D < 0 $$: $$ 4(1 - p^{2}) < 0 $$ Dividing both sides by 4 (which does not change the inequality since 4 is positive): $$ 1 - p^{2} < 0 $$ Rearranging gives: $$ p^{2} > 1 $$ 4. **Finding the values of $$ p $$**: Taking the square root of both sides, we find: $$ |p| > 1 $$ This means: $$ p < -1 \quad \text{or} \quad p > 1 $$ Thus, the values of $$ p $$ for which the roots of the equation are non-real are $$ p < -1 $$ or $$ p > 1 $$. The roots of the equation $$ (x+5)^{2}=1-p^{2} $$ are non-real when $$ p < -1 $$ or $$ p > 1 $$.

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Calculate the values of for which the roots of the equation are non-rea

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Given: Calculate the values of for which the roots of the equation are non-rea