Discover new information and get your questions answered with IDNLearn.com. Ask anything and receive well-informed answers from our community of experienced professionals.
Sagot :
Let's solve each of the quadratic equations step by step and find the roots correct to 2 decimal places.
### Equation (a): [tex]\(2x^2 + 4x - 8 = 0\)[/tex]
1. Identify coefficients: [tex]\(a = 2\)[/tex], [tex]\(b = 4\)[/tex], [tex]\(c = -8\)[/tex].
2. Discriminant: [tex]\(D = b^2 - 4ac = 4^2 - 4 \cdot 2 \cdot (-8) = 16 + 64 = 80\)[/tex].
3. [tex]\(D\)[/tex] is positive, so we have two real roots.
4. Roots:
- [tex]\(x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-4 + \sqrt{80}}{4} \approx 1.24\)[/tex]
- [tex]\(x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-4 - \sqrt{80}}{4} \approx -3.24\)[/tex]
So, the solutions for equation (a) are [tex]\(1.24\)[/tex] and [tex]\(-3.24\)[/tex].
### Equation (b): [tex]\(6x^2 - 3x - 2 = 0\)[/tex]
1. Identify coefficients: [tex]\(a = 6\)[/tex], [tex]\(b = -3\)[/tex], [tex]\(c = -2\)[/tex].
2. Discriminant: [tex]\(D = b^2 - 4ac = (-3)^2 - 4 \cdot 6 \cdot (-2) = 9 + 48 = 57\)[/tex].
3. [tex]\(D\)[/tex] is positive, so we have two real roots.
4. Roots:
- [tex]\(x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{57}}{12} \approx 0.88\)[/tex]
- [tex]\(x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{57}}{12} \approx -0.38\)[/tex]
So, the solutions for equation (b) are [tex]\(0.88\)[/tex] and [tex]\(-0.38\)[/tex].
### Equation (c): [tex]\(3x^2 + 6x - 10 = 0\)[/tex]
1. Identify coefficients: [tex]\(a = 3\)[/tex], [tex]\(b = 6\)[/tex], [tex]\(c = -10\)[/tex].
2. Discriminant: [tex]\(D = b^2 - 4ac = 6^2 - 4 \cdot 3 \cdot (-10) = 36 + 120 = 156\)[/tex].
3. [tex]\(D\)[/tex] is positive, so we have two real roots.
4. Roots:
- [tex]\(x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-6 + \sqrt{156}}{6} \approx 1.08\)[/tex]
- [tex]\(x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-6 - \sqrt{156}}{6} \approx -3.08\)[/tex]
So, the solutions for equation (c) are [tex]\(1.08\)[/tex] and [tex]\(-3.08\)[/tex].
### Equation (d): [tex]\(5x^2 - 15x - 4 = 0\)[/tex]
1. Identify coefficients: [tex]\(a = 5\)[/tex], [tex]\(b = -15\)[/tex], [tex]\(c = -4\)[/tex].
2. Discriminant: [tex]\(D = b^2 - 4ac = (-15)^2 - 4 \cdot 5 \cdot (-4) = 225 + 80 = 305\)[/tex].
3. [tex]\(D\)[/tex] is positive, so we have two real roots.
4. Roots:
- [tex]\(x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{15 + \sqrt{305}}{10} \approx 3.25\)[/tex]
- [tex]\(x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{15 - \sqrt{305}}{10} \approx -0.25\)[/tex]
So, the solutions for equation (d) are [tex]\(3.25\)[/tex] and [tex]\(-0.25\)[/tex].
### Equation (e): [tex]\(4x^2 + 6x - 5 = 0\)[/tex]
1. Identify coefficients: [tex]\(a = 4\)[/tex], [tex]\(b = 6\)[/tex], [tex]\(c = -5\)[/tex].
2. Discriminant: [tex]\(D = b^2 - 4ac = 6^2 - 4 \cdot 4 \cdot (-5) = 36 + 80 = 116\)[/tex].
3. [tex]\(D\)[/tex] is positive, so we have two real roots.
4. Roots:
- [tex]\(x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-6 + \sqrt{116}}{8} \approx 0.6\)[/tex]
- [tex]\(x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-6 - \sqrt{116}}{8} \approx -2.1\)[/tex]
So, the solutions for equation (e) are [tex]\(0.6\)[/tex] and [tex]\(-2.1\)[/tex].
To summarize, the roots of the given quadratic equations are:
- For [tex]\(2x^2 + 4x - 8 = 0\)[/tex]: [tex]\(x_1 \approx 1.24\)[/tex], [tex]\(x_2 \approx -3.24\)[/tex]
- For [tex]\(6x^2 - 3x - 2 = 0\)[/tex]: [tex]\(x_1 \approx 0.88\)[/tex], [tex]\(x_2 \approx -0.38\)[/tex]
- For [tex]\(3x^2 + 6x - 10 = 0\)[/tex]: [tex]\(x_1 \approx 1.08\)[/tex], [tex]\(x_2 \approx -3.08\)[/tex]
- For [tex]\(5x^2 - 15x - 4 = 0\)[/tex]: [tex]\(x_1 \approx 3.25\)[/tex], [tex]\(x_2 \approx -0.25\)[/tex]
- For [tex]\(4x^2 + 6x - 5 = 0\)[/tex]: [tex]\(x_1 \approx 0.6\)[/tex], [tex]\(x_2 \approx -2.1\)[/tex]
### Equation (a): [tex]\(2x^2 + 4x - 8 = 0\)[/tex]
1. Identify coefficients: [tex]\(a = 2\)[/tex], [tex]\(b = 4\)[/tex], [tex]\(c = -8\)[/tex].
2. Discriminant: [tex]\(D = b^2 - 4ac = 4^2 - 4 \cdot 2 \cdot (-8) = 16 + 64 = 80\)[/tex].
3. [tex]\(D\)[/tex] is positive, so we have two real roots.
4. Roots:
- [tex]\(x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-4 + \sqrt{80}}{4} \approx 1.24\)[/tex]
- [tex]\(x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-4 - \sqrt{80}}{4} \approx -3.24\)[/tex]
So, the solutions for equation (a) are [tex]\(1.24\)[/tex] and [tex]\(-3.24\)[/tex].
### Equation (b): [tex]\(6x^2 - 3x - 2 = 0\)[/tex]
1. Identify coefficients: [tex]\(a = 6\)[/tex], [tex]\(b = -3\)[/tex], [tex]\(c = -2\)[/tex].
2. Discriminant: [tex]\(D = b^2 - 4ac = (-3)^2 - 4 \cdot 6 \cdot (-2) = 9 + 48 = 57\)[/tex].
3. [tex]\(D\)[/tex] is positive, so we have two real roots.
4. Roots:
- [tex]\(x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{57}}{12} \approx 0.88\)[/tex]
- [tex]\(x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{57}}{12} \approx -0.38\)[/tex]
So, the solutions for equation (b) are [tex]\(0.88\)[/tex] and [tex]\(-0.38\)[/tex].
### Equation (c): [tex]\(3x^2 + 6x - 10 = 0\)[/tex]
1. Identify coefficients: [tex]\(a = 3\)[/tex], [tex]\(b = 6\)[/tex], [tex]\(c = -10\)[/tex].
2. Discriminant: [tex]\(D = b^2 - 4ac = 6^2 - 4 \cdot 3 \cdot (-10) = 36 + 120 = 156\)[/tex].
3. [tex]\(D\)[/tex] is positive, so we have two real roots.
4. Roots:
- [tex]\(x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-6 + \sqrt{156}}{6} \approx 1.08\)[/tex]
- [tex]\(x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-6 - \sqrt{156}}{6} \approx -3.08\)[/tex]
So, the solutions for equation (c) are [tex]\(1.08\)[/tex] and [tex]\(-3.08\)[/tex].
### Equation (d): [tex]\(5x^2 - 15x - 4 = 0\)[/tex]
1. Identify coefficients: [tex]\(a = 5\)[/tex], [tex]\(b = -15\)[/tex], [tex]\(c = -4\)[/tex].
2. Discriminant: [tex]\(D = b^2 - 4ac = (-15)^2 - 4 \cdot 5 \cdot (-4) = 225 + 80 = 305\)[/tex].
3. [tex]\(D\)[/tex] is positive, so we have two real roots.
4. Roots:
- [tex]\(x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{15 + \sqrt{305}}{10} \approx 3.25\)[/tex]
- [tex]\(x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{15 - \sqrt{305}}{10} \approx -0.25\)[/tex]
So, the solutions for equation (d) are [tex]\(3.25\)[/tex] and [tex]\(-0.25\)[/tex].
### Equation (e): [tex]\(4x^2 + 6x - 5 = 0\)[/tex]
1. Identify coefficients: [tex]\(a = 4\)[/tex], [tex]\(b = 6\)[/tex], [tex]\(c = -5\)[/tex].
2. Discriminant: [tex]\(D = b^2 - 4ac = 6^2 - 4 \cdot 4 \cdot (-5) = 36 + 80 = 116\)[/tex].
3. [tex]\(D\)[/tex] is positive, so we have two real roots.
4. Roots:
- [tex]\(x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-6 + \sqrt{116}}{8} \approx 0.6\)[/tex]
- [tex]\(x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-6 - \sqrt{116}}{8} \approx -2.1\)[/tex]
So, the solutions for equation (e) are [tex]\(0.6\)[/tex] and [tex]\(-2.1\)[/tex].
To summarize, the roots of the given quadratic equations are:
- For [tex]\(2x^2 + 4x - 8 = 0\)[/tex]: [tex]\(x_1 \approx 1.24\)[/tex], [tex]\(x_2 \approx -3.24\)[/tex]
- For [tex]\(6x^2 - 3x - 2 = 0\)[/tex]: [tex]\(x_1 \approx 0.88\)[/tex], [tex]\(x_2 \approx -0.38\)[/tex]
- For [tex]\(3x^2 + 6x - 10 = 0\)[/tex]: [tex]\(x_1 \approx 1.08\)[/tex], [tex]\(x_2 \approx -3.08\)[/tex]
- For [tex]\(5x^2 - 15x - 4 = 0\)[/tex]: [tex]\(x_1 \approx 3.25\)[/tex], [tex]\(x_2 \approx -0.25\)[/tex]
- For [tex]\(4x^2 + 6x - 5 = 0\)[/tex]: [tex]\(x_1 \approx 0.6\)[/tex], [tex]\(x_2 \approx -2.1\)[/tex]
Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. Thank you for visiting IDNLearn.com. We’re here to provide clear and concise answers, so visit us again soon.
