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\( \begin{array}{l} 2 x^{2}+3 x-1=0 \\ 2 x^{2}-30 x+25=0 \end{array} {} x^{2}-23 x+120=0 {} 16 x^{2}-24 x+9=0 \)

Ask by Savage Johnson. in Pakistan
Jan 24,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The solutions to the equations are: 1. \( x = \frac{-3 \pm \sqrt{17}}{4} \) (approximately \( x \approx 0.2808 \) and \( x \approx -1.7808 \)) 2. \( x = \frac{15 \pm 5\sqrt{7}}{2} \) (approximately \( x \approx 14.1146 \) and \( x \approx 0.8854 \)) 3. \( x = 15 \) and \( x = 8 \) 4. \( x = \frac{3}{4} \)

Solution

Sure! Let's solve each of the quadratic equations you've provided. We'll use the quadratic formula for each, which is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a \), \( b \), and \( c \) are coefficients from the quadratic equation \( ax^2 + bx + c = 0 \). --- ### 1. \( 2x^{2} + 3x - 1 = 0 \) **Coefficients:** - \( a = 2 \) - \( b = 3 \) - \( c = -1 \) **Discriminant (\( D \)):** \[ D = b^2 - 4ac = 3^2 - 4(2)(-1) = 9 + 8 = 17 \] **Solutions:** \[ x = \frac{-3 \pm \sqrt{17}}{2 \times 2} = \frac{-3 \pm \sqrt{17}}{4} \] **Approximate Values:** \[ x \approx \frac{-3 + 4.1231}{4} \approx \frac{1.1231}{4} \approx 0.2808 \] \[ x \approx \frac{-3 - 4.1231}{4} \approx \frac{-7.1231}{4} \approx -1.7808 \] --- ### 2. \( 2x^{2} - 30x + 25 = 0 \) **Coefficients:** - \( a = 2 \) - \( b = -30 \) - \( c = 25 \) **Discriminant (\( D \)):** \[ D = (-30)^2 - 4(2)(25) = 900 - 200 = 700 \] **Solutions:** \[ x = \frac{30 \pm \sqrt{700}}{2 \times 2} = \frac{30 \pm 10\sqrt{7}}{4} = \frac{15 \pm 5\sqrt{7}}{2} \] **Approximate Values:** \[ \sqrt{7} \approx 2.6458 \] \[ x \approx \frac{30 + 26.4583}{4} = \frac{56.4583}{4} \approx 14.1146 \] \[ x \approx \frac{30 - 26.4583}{4} = \frac{3.5417}{4} \approx 0.8854 \] --- ### 3. \( x^{2} - 23x + 120 = 0 \) **Coefficients:** - \( a = 1 \) - \( b = -23 \) - \( c = 120 \) **Discriminant (\( D \)):** \[ D = (-23)^2 - 4(1)(120) = 529 - 480 = 49 \] **Solutions:** \[ x = \frac{23 \pm \sqrt{49}}{2 \times 1} = \frac{23 \pm 7}{2} \] **Values:** \[ x = \frac{23 + 7}{2} = \frac{30}{2} = 15 \] \[ x = \frac{23 - 7}{2} = \frac{16}{2} = 8 \] --- ### 4. \( 16x^{2} - 24x + 9 = 0 \) **Coefficients:** - \( a = 16 \) - \( b = -24 \) - \( c = 9 \) **Discriminant (\( D \)):** \[ D = (-24)^2 - 4(16)(9) = 576 - 576 = 0 \] Since the discriminant is zero, there is **one real repeated root**. **Solution:** \[ x = \frac{24}{2 \times 16} = \frac{24}{32} = \frac{3}{4} \] **Alternatively, Factoring:** \[ 16x^2 - 24x + 9 = (4x - 3)^2 = 0 \implies 4x - 3 = 0 \implies x = \frac{3}{4} \] --- ### Summary of Solutions 1. **\( 2x^{2} + 3x - 1 = 0 \)** - \( x = \frac{-3 \pm \sqrt{17}}{4} \) - Approximately \( x \approx 0.2808 \) and \( x \approx -1.7808 \) 2. **\( 2x^{2} - 30x + 25 = 0 \)** - \( x = \frac{15 \pm 5\sqrt{7}}{2} \) - Approximately \( x \approx 14.1146 \) and \( x \approx 0.8854 \) 3. **\( x^{2} - 23x + 120 = 0 \)** - \( x = 15 \) and \( x = 8 \) 4. **\( 16x^{2} - 24x + 9 = 0 \)** - \( x = \frac{3}{4} \) If you have any further questions or need additional explanations for any of these solutions, feel free to ask!

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Beyond the Answer

Did you know that the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), is like a magical spell for solving any quadratic equation? This formula was first recorded in ancient Babylon around 2000 BC! Even though they didn’t have the notation we use today, they were already solving for \( x \) using geometric methods. It’s amazing how math transcends time and evolves through cultures! In real life, quadratic equations pop up in various scenarios, such as calculating the trajectory of projectiles or determining areas in land surveying. For instance, if you’re throwing a ball, its height at any time can be modeled by a quadratic equation. So next time you’re at the park tossing a ball, just remember that you’re applying some algebraic principles from centuries of mathematical journey!

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