Let's solve each of the problems step-by-step.
1.1 Solve for \( x \):
1.1.1 \( 5x^2 + 2x - 6 = 0 \)
We can use the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \), where \( a = 5 \), \( b = 2 \), and \( c = -6 \).
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Calculate the discriminant: \[ D = b^2 - 4ac = 2^2 - 4(5)(-6) = 4 + 120 = 124 \]
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Calculate \( x \): \[ x = \frac{{-2 \pm \sqrt{124}}}{{2 \cdot 5}} = \frac{{-2 \pm 2\sqrt{31}}}{10} = \frac{{-1 \pm \sqrt{31}}}{5} \]
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Find the approximate values:
- \( \sqrt{31} \approx 5.57 \), so: \[ x_1 \approx \frac{{-1 + 5.57}}{5} \approx \frac{4.57}{5} \approx 0.914 \quad \text{(to TWO decimal places, } 0.91\text{)} \] \[ x_2 \approx \frac{{-1 - 5.57}}{5} \approx \frac{-6.57}{5} \approx -1.314 \quad \text{(to TWO decimal places, } -1.31\text{)} \]
Solutions: \( x \approx 0.91 \) and \( x \approx -1.31 \)
1.1.2 \( (x + 4)(x - 1) = 0 \)
Set each factor equal to zero:
- \( x + 4 = 0 \) โ \( x = -4 \)
- \( x - 1 = 0 \) โ \( x = 1 \)
Solutions: \( x = -4 \) and \( x = 1 \)
1.1.3 \( (2x - 1)(x + 3) \geq -3 \)
First, rearrange the inequality: \[ (2x - 1)(x + 3) + 3 \geq 0 \] Let's expand: \[ 2x^2 + 6x - x - 3 + 3 \geq 0 \] This simplifies to: \[ 2x^2 + 5x \geq 0 \] Factoring out \( x \): \[ x(2x + 5) \geq 0 \]
Determine the critical points: \( x = 0 \) and \( 2x + 5 = 0 \) gives \( x = -\frac{5}{2} \).
Test intervals:
- For \( x < -\frac{5}{2} \) (e.g., \( x = -3 \)): \( -3 \cdot (-6) > 0 \) โ True
- For \( -\frac{5}{2} < x < 0 \) (e.g., \( x = -1 \)): \( -1 \cdot 3 < 0 \) โ False
- For \( x > 0 \) (e.g., \( x = 1 \)): \( 1 \cdot 7 > 0 \) โ True
Solutions: \( x \leq -\frac{5}{2} \) or \( x \geq 0 \)
1.1.4 \( 5 - x + 1 = -x \)
Rearranging gives: \[ 6 - x = -x \] \[ 6 = 0 \quad \text{(which is not possible)} \]
No solution.
1.2 Solve for \( x \) and \( y \) simultaneously:
Given the equations:
- \( 2x + y = 1 \) (Equation 1)
- \( 2x^2 - xy + y^2 = 4 \) (Equation 2)
From Equation 1: \( y = 1 - 2x \)
Substituting into Equation 2: \[ 2x^2 - x(1 - 2x) + (1 - 2x)^2 = 4 \] Expanding it yields: \[ 2x^2 - x + 2x^2 + 1 - 4x + 4x^2 = 4 \] Combine like terms: \[ 8x^2 - 5x + 1 = 4 \] \[ 8x^2 - 5x - 3 = 0 \] Apply the quadratic formula: \[ x = \frac{{-(-5) \pm \sqrt{{(-5)^2 - 4 \cdot 8 \cdot (-3)}}}}{{2 \cdot 8}} = \frac{{5 \pm \sqrt{{25 + 96}}}}{{16}} = \frac{{5 \pm \sqrt{121}}}{16} = \frac{{5 \pm 11}}{16} \] Calculating roots:
- \( x_1 = \frac{16}{16} = 1 \)
- \( x_2 = \frac{-6}{16} = -\frac{3}{8} \)
Finding \( y \): For \( x = 1 \): \[ y = 1 - 2(1) = -1 \] For \( x = -\frac{3}{8} \): \[ y = 1 - 2\left(-\frac{3}{8}\right) = 1 + \frac{6}{8} = \frac{14}{8} = \frac{7}{4} \]
Solutions: \( (x, y) = (1, -1) \) and \( \left(-\frac{3}{8}, \frac{7}{4}\right) \)
1.3 The roots of a quadratic equation:
Given: \[ x = \frac{{-4 \pm (k+1)(3-k)}}{2} \]
1.3.1 If \( k = 2 \), determine the nature of its roots.
Substituting \( k = 2 \): \[ x = \frac{{-4 \pm (2+1)(3-2)}}{2} = \frac{{-4 \pm 3}}{2} \] Calculating the roots: \[ x_1 = \frac{{-4 + 3}}{2} = \frac{{-1}}{2} = -0.5 \] \[ x_2 = \frac{{-4 - 3}}{2} = \frac{{-7}}{2} = -3.5 \]
Both roots exist and are real.
1.3.2 Determine the value(s) of \( k \) for which the roots are non-real.
The roots are non-real if the term \( (k + 1)(3 - k) = 0 \) becomes non-negative. Setting the product to zero:
- \( k + 1 = 0 \) โ \( k = -1 \)
- \( 3 - k = 0 \) โ \( k = 3 \)
The roots will be real when the quadratic is non-negative: \((k + 1)(3 - k) < 0\)
- This occurs between the roots \( k = -1 \) and \( k = 3 \):
Thus, \( k < -1 \) or \( k > 3 \) leads to non-real roots.
In summary:
- \( k < -1 \) or \( k > 3 \): non-real roots.
- \( k = -1 \) or \( k = 3 \): roots are real.
Final Summary of Solutions: 1.1.1: \( x \approx 0.91, -1.31 \) 1.1.2: \( x = -4, 1 \) 1.1.3: \( x \leq -\frac{5}{2} \) or \( x \geq 0 \) 1.1.4: No solution. 1.2: \( (x, y) = (1, -1) \) and \( \left(-\frac{3}{8}, \frac{7}{4}\right) \) 1.3.1: Roots are real. 1.3.2: Non-real for \( k < -1 \) or \( k > 3 \).
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