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1. Without soluing each of the following equations, discues the hat ure of the roots: a) \( x^{2}+x+1=0 \). b) \( x^{2}=2(x+1) \) c) \( x^{2}=4 x \) c) \( -2 x^{2}-16 x-32=0 \) c) \( (x-2)(2 x-1)=5 \). Howrewook 12 Marchzoz5. 1) Show that the roots of the: equation \( x(x-3 m)=-5 m^{2} \). are non-real if \( m \neq 0 \). 2) Show that the roots of the equation \( x^{2}+(p-2) x+(1-p)=0 \) are rational for any real value of \( p \).

Ask by Simmons Norton. in South Africa
Mar 13,2025

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- a) The equation \( x^{2}+x+1=0 \) has non-real (complex) roots. - b) The equation \( x^{2}=2(x+1) \) has real and distinct roots. - c) The equation \( x^{2}=4x \) has real and distinct roots. - d) The equation \( -2 x^{2}-16 x-32=0 \) has real and equal roots. - e) The equation \( (x-2)(2x-1)=5 \) has real and distinct roots. For the specific equations: 1) The roots of \( x(x-3m)=-5m^{2} \) are non-real if \( m \neq 0 \). 2) The roots of \( x^{2}+(p-2)x+(1-p)=0 \) are rational for any real value of \( p \).

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Bonus Knowledge

Let’s dive into some roots analysis! For the equation \( x^{2}+x+1=0 \), you can determine that the discriminant \( b^2 - 4ac \) is negative (specifically, \( 1 - 4 \)), indicating that the roots are complex and non-real. In contrast, \( x^{2}=2(x+1) \) simplifies to \( x^2 - 2x - 2 = 0 \), and its discriminant is positive (8), suggesting two distinct real roots! For the second batch, the first equation \( x(x-3m)=-5m^2 \) can be rearranged to a standard quadratic form. Here, since \( m \neq 0 \), this results in a negative discriminant, thus confirming complex roots. In the case of \( x^{2}+(p-2)x+(1-p)=0 \), the discriminant simplifies to a perfect square, revealing that roots are indeed rational regardless of the value of \( p \) since the expression is always a square when reconfigured. Now, onto some captivating insights! The equation \( x^2 + x + 1 = 0 \) might remind you of its roots, which are also connected to the cube roots of unity! These mystical roots have applications even in solving polynomial equations in advanced mathematics, navigating through complex planes. As for rational roots, the discriminant often reveals much about the nature of the roots promptly. If you’re interested in experimenting further, using the Rational Root Theorem could be a fun exploration for identifying potential rational candidates in any polynomial equation!

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