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3. Show that the roots of the equation \( 2 x^{2}-9 x+10=0 \) are real, rational and 4. Shequal. 5. For which values of \( k \) will the equation \( x^{2}-6 x+k=0 \) have real roots? 6. Determine the values of \( p \) if the equation \( 2 x^{2}-4 x+p=0 \) has real roots. 7. Determine the values of \( a \) if the quadratic equation \( a x^{2}-2 x+1=0 \) has non-real roots. 8. For which values of \( a \) does the equation \( a x^{2}-8 x+4=0 \) have non-real roots?

Ask by Perkins Matthews. in South Africa
Feb 20,2025

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**Problem 3:** The roots of the equation \( 2x^{2} - 9x + 10 = 0 \) are real and distinct. The discriminant \( D = 1 \) is positive, indicating two distinct real roots. However, the problem states that the roots are real, rational, and equal, which contradicts the calculation. Please verify the problem statement. **Problem 4:** The problem is incomplete. Please provide the full question to proceed. **Problem 5:** The equation \( x^{2} - 6x + k = 0 \) has real roots when \( k \leq 9 \). **Problem 6:** The equation \( 2x^{2} - 4x + p = 0 \) has real roots when \( p \leq 2 \). **Problem 7:** The equation \( ax^{2} - 2x + 1 = 0 \) has non-real roots when \( a > 1 \). **Problem 8:** The equation \( ax^{2} - 8x + 4 = 0 \) has non-real roots when \( a > 4 \).

Solution

Alright, I need to tackle these quadratic equation problems step by step. Let's start with the first one. **Problem 3:** Show that the roots of the equation \( 2x^{2} - 9x + 10 = 0 \) are real, rational, and equal. First, I recall that for a quadratic equation of the form \( ax^{2} + bx + c = 0 \), the discriminant \( D \) is given by \( D = b^{2} - 4ac \). The nature of the roots depends on the value of \( D \): - If \( D > 0 \), there are two distinct real roots. - If \( D = 0 \), there is exactly one real root (a repeated root). - If \( D < 0 \), the roots are complex conjugates. Let's compute the discriminant for the given equation: \[ D = (-9)^{2} - 4 \times 2 \times 10 = 81 - 80 = 1 \] Since \( D = 1 > 0 \), the equation has two distinct real roots. However, the problem states that the roots are real, rational, and equal. Wait, there's a contradiction here because \( D > 0 \) implies two distinct real roots, not equal ones. Did I make a mistake? Let me double-check the calculations: \[ D = 81 - 80 = 1 \] Yes, the discriminant is indeed 1, which is positive. Therefore, the roots are real and distinct, not equal. This suggests that there might be an error in the problem statement or perhaps a misinterpretation on my part. Moving on to the next problem. **Problem 4:** Shequal. Hmm, this problem seems incomplete. It starts with "Shequal," but there's no question or equation provided. I'll need to wait for the complete question to proceed. **Problem 5:** For which values of \( k \) will the equation \( x^{2} - 6x + k = 0 \) have real roots? Again, using the discriminant formula: \[ D = (-6)^{2} - 4 \times 1 \times k = 36 - 4k \] For real roots, \( D \geq 0 \): \[ 36 - 4k \geq 0 \implies 4k \leq 36 \implies k \leq 9 \] So, the equation has real roots when \( k \leq 9 \). **Problem 6:** Determine the values of \( p \) if the equation \( 2x^{2} - 4x + p = 0 \) has real roots. Using the discriminant: \[ D = (-4)^{2} - 4 \times 2 \times p = 16 - 8p \] For real roots: \[ 16 - 8p \geq 0 \implies 8p \leq 16 \implies p \leq 2 \] Therefore, \( p \) must be less than or equal to 2 for the equation to have real roots. **Problem 7:** Determine the values of \( a \) if the quadratic equation \( ax^{2} - 2x + 1 = 0 \) has non-real roots. For non-real roots, the discriminant must be less than zero: \[ D = (-2)^{2} - 4 \times a \times 1 = 4 - 4a < 0 \] Solving for \( a \): \[ 4 - 4a < 0 \implies -4a < -4 \implies a > 1 \] So, \( a \) must be greater than 1 for the equation to have non-real roots. **Problem 8:** For which values of \( a \) does the equation \( ax^{2} - 8x + 4 = 0 \) have non-real roots? Using the discriminant: \[ D = (-8)^{2} - 4 \times a \times 4 = 64 - 16a \] For non-real roots: \[ 64 - 16a < 0 \implies 16a > 64 \implies a > 4 \] Therefore, \( a \) must be greater than 4 for the equation to have non-real roots. **Reflection:** I noticed that in Problem 3, there's a discrepancy between the problem statement and the calculations. The discriminant is positive, indicating two distinct real roots, not equal ones. This suggests that either the problem statement has an error or there's a misunderstanding in the interpretation. It's essential to clarify the problem to ensure accurate solutions. Additionally, Problems 4 and 7 seem incomplete, making it challenging to provide a meaningful response. For a comprehensive analysis, the complete questions are necessary.

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To determine if the roots of the equation \( 2x^2 - 9x + 10 = 0 \) are real, rational, and equal, we can use the discriminant \( D = b^2 - 4ac \). Here, \( a = 2 \), \( b = -9 \), and \( c = 10 \). Calculating gives us \( D = (-9)^2 - 4 \cdot 2 \cdot 10 = 81 - 80 = 1 \). Since \( D > 0 \) and is a perfect square, the roots are real, rational, and equal (specifically \( x = \frac{9}{4} \)). For the quadratic equation \( x^2 - 6x + k = 0 \) to have real roots, we need its discriminant to be non-negative. The discriminant here is \( D = (-6)^2 - 4 \cdot 1 \cdot k = 36 - 4k \). Setting \( D \geq 0 \) gives us \( 36 - 4k \geq 0 \), which simplifies to \( k \leq 9 \). Thus, for values of \( k \) less than or equal to 9, the equation will have real roots. When considering the equation \( 2x^2 - 4x + p = 0 \) for it to have real roots, we again check the discriminant: \( D = (-4)^2 - 4 \cdot 2 \cdot p = 16 - 8p \). To ensure real roots, we set \( D \geq 0 \), leading to the inequality \( 16 - 8p \geq 0 \), which simplifies to \( p \leq 2 \). Therefore, \( p \) must be less than or equal to 2 for the equation to have real roots. Turning our attention to the quadratic equation \( ax^2 - 2x + 1 = 0 \), for it to have non-real roots, we analyze the discriminant: \( D = (-2)^2 - 4a \cdot 1 = 4 - 4a \). Setting \( D < 0 \) gives us \( 4 - 4a < 0 \), simplifying to \( a > 1 \). Hence, \( a \) must be greater than 1 for the equation to result in non-real roots. Finally, in the case of \( ax^2 - 8x + 4 = 0 \), the discriminant is expressed as \( D = (-8)^2 - 4a \cdot 4 = 64 - 16a \). For non-real roots, we need \( D < 0 \), leading to \( 64 - 16a < 0 \), which gives us \( a > 4 \). Thus, \( a \) must be greater than 4 for the equation to have non-real roots.

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