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\( 2.12 x^{2}+2 x+1=0 \) solve for \( x \)

Ask by Warren Malone. in South Africa
Mar 14,2025

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The solutions are \( x = \frac{-2 \pm i\sqrt{4.48}}{4.24} \) or approximately \( x \approx -0.4717 \pm 0.5i \).

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The Deep Dive

To solve the equation \( 2.12 x^{2}+2 x+1=0 \), you can apply the quadratic formula: \[ x = \frac{{-b \pm \sqrt{{b^{2}-4ac}}}}{2a} \] Here, \( a = 2.12 \), \( b = 2 \), and \( c = 1 \). Plugging these values in, we calculate: \[ b^{2} - 4ac = 2^{2} - 4(2.12)(1) = 4 - 8.48 = -4.48 \] Since the discriminant is negative, this equation has no real solutions, but two complex solutions. So we proceed: \[ x = \frac{{-2 \pm \sqrt{{-4.48}}}}{2(2.12)} = \frac{{-2 \pm 2.11i}}{4.24} = \frac{{-1 \pm 1.05i}}{2.12} \] The solutions are: \[ x = \frac{-1 + 1.05i}{2.12}, \quad x = \frac{-1 - 1.05i}{2.12} \] This showcases the fascinating world of complex numbers! Additionally, if you ever encounter equations largely like this, remember that sometimes rechecking whether a factorable form exists or adjusting coefficients can simplify the process!

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