Pregunta
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(a) Discuss the nature of the roots of the following quadratic equations, without solving the equations: (1) \( x^{2}-3 x-10=0 \) (2) \( x^{2}-2 x+1=0 \) (3) \( x^{2}-5 x-7=0 \) (4) \( x^{2}-x+3=0 \) (5) \( 2 x^{2}+6=-7 x \) (6) \( x+\frac{6}{x}=4 \) (77) \( (x+2)(x-1)=4 x \) (8) \( x^{2}+p^{2}=2 p x \) (9)) \( k x^{2}+(4 k-1) x=2-4 k \) (b) For which value(s) of \( m \) will the following expressions have real roots? (1) \( x^{2}+6 x+m=0 \) (2) \( 3 x^{2}=2 m x-3 \) (3) \( m x^{2}+m x+m=3 \) (c) For which value(s) of \( m \) will the following expressions have equal roots? (11) \( x^{2}+2 m=8 x \) (2) \( x^{2}+m x+9=0 \) ((3)) \( x^{2}-m x+2 x+1=0 \) (d) For which value(s) of \( m \) will the following expressions have non-real roots? (1)) \( x^{2}-6 x=m \) (2) \( x^{2}+m x+4=3 x \) (3) \( \frac{x^{2}-5 x+5}{2 x^{2}-x+1}=\frac{1}{m} \) Show that the equation \( x^{2}+(k-2) x-k=0 \) has real roots for all real values of \( k \).

Ask by Chadwick Phillips. in South Africa
Feb 19,2025

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**Part (a): Nature of the Roots** 1. **\( x^{2}-3x-10=0 \)**: Two distinct real roots. 2. **\( x^{2}-2x+1=0 \)**: One real root. 3. **\( x^{2}-5x-7=0 \)**: Two distinct real roots. 4. **\( x^{2}-x+3=0 \)**: No real roots. 5. **\( 2x^{2}+6=-7x \)**: Two distinct real roots. 6. **\( x+\frac{6}{x}=4 \)**: No real roots. 7. **\( (x+2)(x-1)=4x \)**: Two distinct real roots. 8. **\( x^{2}+p^{2}=2px \)**: One real root. 9. **\( kx^{2}+(4k-1)x=2-4k \)**: Two distinct real roots. **Part (b): Values of \( m \) for Real Roots** 1. **\( x^{2}+6x+m=0 \)**: \( m \leq 9 \). 2. **\( 3x^{2}=2mx-3 \)**: \( m \leq -3 \) or \( m \geq 3 \). 3. **\( mx^{2}+mx+m=3 \)**: \( m \leq 0 \) or \( m \geq 4 \). **Part (c): Values of \( m \) for Equal Roots** 1. **\( x^{2}+2m=8x \)**: \( m = 0 \) or \( m = 4 \). 2. **\( x^{2}+mx+9=0 \)**: \( m = \pm 6 \). 3. **\( x^{2}-mx+2x+1=0 \)**: \( m = 3 \) or \( m = -1 \). **Part (d): Values of \( m \) for Non-Real Roots** 1. **\( x^{2}-6x=m \)**: \( m > 9 \). 2. **\( x^{2}+mx+4=3x \)**: \( m < -6 \) or \( m > 6 \). 3. **\( \frac{x^{2}-5x+5}{2x^{2}-x+1}=\frac{1}{m} \)**: \( m \neq 0 \). **Additional Information:** - The equation \( x^{2}+(k-2)x-k=0 \) has real roots for all real values of \( k \) because the discriminant \( D = (k-2)^2 - 4(1)(-k) = k^2 - 4k + 4 + 4k = k^2 + 4 \) is always positive.

Solución

To analyze the nature of the roots of the given quadratic equations, we will use the discriminant \( D \) of a quadratic equation of the form \( ax^2 + bx + c = 0 \). The discriminant is given by: \[ D = b^2 - 4ac \] The nature of the roots can be determined as follows: - If \( D > 0 \), the equation has two distinct real roots. - If \( D = 0 \), the equation has exactly one real root (a repeated root). - If \( D < 0 \), the equation has no real roots (the roots are complex). ### Part (a): Nature of the Roots 1. **For \( x^{2}-3x-10=0 \)**: - Here, \( a = 1 \), \( b = -3 \), \( c = -10 \). - Calculate \( D = (-3)^2 - 4(1)(-10) = 9 + 40 = 49 \). - Since \( D > 0 \), there are two distinct real roots. 2. **For \( x^{2}-2x+1=0 \)**: - Here, \( a = 1 \), \( b = -2 \), \( c = 1 \). - Calculate \( D = (-2)^2 - 4(1)(1) = 4 - 4 = 0 \). - Since \( D = 0 \), there is exactly one real root. 3. **For \( x^{2}-5x-7=0 \)**: - Here, \( a = 1 \), \( b = -5 \), \( c = -7 \). - Calculate \( D = (-5)^2 - 4(1)(-7) = 25 + 28 = 53 \). - Since \( D > 0 \), there are two distinct real roots. 4. **For \( x^{2}-x+3=0 \)**: - Here, \( a = 1 \), \( b = -1 \), \( c = 3 \). - Calculate \( D = (-1)^2 - 4(1)(3) = 1 - 12 = -11 \). - Since \( D < 0 \), there are no real roots. 5. **For \( 2x^{2}+6=-7x \)**: - Rearranging gives \( 2x^{2} + 7x + 6 = 0 \). - Here, \( a = 2 \), \( b = 7 \), \( c = 6 \). - Calculate \( D = (7)^2 - 4(2)(6) = 49 - 48 = 1 \). - Since \( D > 0 \), there are two distinct real roots. 6. **For \( x+\frac{6}{x}=4 \)**: - Multiplying through by \( x \) gives \( x^2 - 4x + 6 = 0 \). - Here, \( a = 1 \), \( b = -4 \), \( c = 6 \). - Calculate \( D = (-4)^2 - 4(1)(6) = 16 - 24 = -8 \). - Since \( D < 0 \), there are no real roots. 7. **For \( (x+2)(x-1)=4x \)**: - Expanding gives \( x^2 + x - 2 = 4x \) or \( x^2 - 3x - 2 = 0 \). - Here, \( a = 1 \), \( b = -3 \), \( c = -2 \). - Calculate \( D = (-3)^2 - 4(1)(-2) = 9 + 8 = 17 \). - Since \( D > 0 \), there are two distinct real roots. 8. **For \( x^{2}+p^{2}=2px \)**: - Rearranging gives \( x^2 - 2px + p^2 = 0 \). - Here, \( a = 1 \), \( b = -2p \), \( c = p^2 \). - Calculate \( D = (-2p)^2 - 4(1)(p^2) = 4p^2 - 4p^2 = 0 \). - Since \( D = 0 \), there is exactly one real root. 9. **For \( kx^{2}+(4k-1)x=2-4k \)**: - Rearranging gives \( kx^{2} + (4k-1)x + (4k-2) = 0 \). - Here, \( a = k \), \( b = 4k - 1 \), \( c = 4k - 2 \). - Calculate \( D = (4k - 1)^2 - 4k(4k - 2) = 16k^2 - 8k + 1 - (16k^2 - 8k) = 1 \). - Since \( D > 0 \), there are two distinct real roots. ### Part (b): Values of \( m \) for Real Roots 1. **For \( x^{2}+6x+m=0 \)**: - \( D = 6^2 - 4(1)(m) = 36 - 4m \geq 0 \). - Thus, \( 36 \geq 4m \) or \( m \leq 9 \). 2. **For \( 3x^{2}=2mx-3 \)**: - Rearranging gives \( 3x^{2} - 2mx + 3 = 0 \). - \( D = (-2m)^2 - 4(3)(3) = 4m^2 - 36 \geq 0 \). - Thus, \( 4m^2 \geq 36 \) or \( m^2 \geq 9 \) leading to \( m \leq -3 \) or \( m \geq 3 \). 3. **For \( mx^{2}+mx+m=3 \)**: - Rearranging gives \( mx^{2} + mx + (m - 3) = 0 \). - \( D = m^2 - 4m(m - 3) = m^2 - 4m^2 + 12m = -3m^2 + 12m \geq 0 \). - Thus, \( -3m(m - 4) \geq 0 \) leading to \( m \leq 0 \) or \( m \geq 4 \). ### Part (c): Values of \( m \) for Equal Roots 1.

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The Deep Dive

For the equation \(x^{2}+(k-2)x-k=0\), we can analyze the nature of the roots by examining the discriminant, which is given by \(\Delta = b^{2} - 4ac\). Here, \(a = 1\), \(b = k-2\), and \(c = -k\). Thus, the discriminant becomes \(\Delta = (k-2)^{2} - 4(1)(-k)= (k-2)^{2} + 4k\). Expanding this gives us \(k^{2} - 4k + 4 + 4k = k^{2} + 4\). Since the square of any real number is non-negative and 4 is always positive, \(\Delta \geq 4\). This implies that the roots are always real for any real value of \(k\). Now, let's talk about some practical situations. Understanding the nature of roots is incredibly useful! Suppose you're working on a design project, and you need to ensure that your quadratic functions representing curves can achieve specific dimensions. Knowing if the roots are real helps you understand if the design can be realized or if you need to adjust your parameters. Additionally, mastering the discriminant can save you from common pitfalls! A frequent mistake is overlooking the possibility of complex roots. Remember the discriminant not only indicates the number of roots but also gives insights into their nature. By getting comfortable with these concepts, you’ll enhance your problem-solving capabilities and avoid falling into the trap of incorrect assumptions about your equations!

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