Example Question - roots
Here are examples of questions we've helped users solve.
Expressing a Number in Exponential Form
<p>Given \( 65^{\frac{4}{5}} = b \cdot \sqrt{a^c} \), we need to express the given equation in terms of \( a \), \( b \), and \( c \).</p> <p>From the equation:</p> <p>Let \( a = 65 \), \( b = 4 \), and \( c = 5 \).</p> <p>Therefore, the values are:</p> <p> \( a = 65 \), \( b = 4 \), \( c = 5 \).</p>
Determining the Nature of Functions
<p>Given the function \( f(x) = x^2 - 5 \).</p> <p>To determine if the function has real roots, we can find the discriminant.</p> <p>The discriminant \( D \) is calculated as \( D = b^2 - 4ac \), where \( a = 1, b = 0, c = -5 \).</p> <p>Thus, \( D = 0^2 - 4 \cdot 1 \cdot (-5) = 20 \).</p> <p>Since \( D > 0 \), the function has two distinct real roots.</p> <p>To find the roots, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \).</p> <p>So, \( x = \frac{-0 \pm \sqrt{20}}{2 \cdot 1} = \frac{\pm 2\sqrt{5}}{2} = \pm \sqrt{5} \).</p> <p>Therefore, the roots are \( x = \sqrt{5} \) and \( x = -\sqrt{5} \).</p>
Quadratic Equation Problem
<p>Para resolver la ecuación cuadrática \( x^2 + 3x - 80 = 0 \), primero buscamos dos números que multiplicados den -80 y sumados den 3. Los números que cumplen con esto son 10 y -8, ya que \( 10 \cdot (-8) = -80 \) y \( 10 + (-8) = 2 \).</p> <p>Entonces, expresamos la ecuación cuadrática como el producto de dos binomios:</p> <p>\[ (x + 10)(x - 8) = 0 \]</p> <p>Luego, igualamos cada binomio a cero y resolvemos para \( x \):</p> <p>\[ x + 10 = 0 \quad \Rightarrow \quad x = -10 \]</p> <p>\[ x - 8 = 0 \quad \Rightarrow \quad x = 8 \]</p> <p>Por lo tanto, las soluciones de la ecuación cuadrática son \( x = -10 \) y \( x = 8 \).</p>
Solving Quadratic Equation using Quadratic Formula
To solve the quadratic equation \(3x^2 + 5x - 2 = 0\), use the quadratic formula: \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\) Here, \(a = 3\), \(b = 5\), and \(c = -2\). Step 1: Calculate the discriminant Discriminant \(D = b^2 - 4ac = 5^2 - 4(3)(-2) = 25 + 24 = 49\) Step 2: Calculate the two roots using the quadratic formula \(x = \frac{{-5 \pm \sqrt{49}}}{{2(3)}}\) Step 3: Compute the two values of \(x\) \(x = \frac{{-5 \pm 7}}{{6}}\) Roots: \(x_1 = \frac{{-5 + 7}}{6} = \frac{2}{6} = \frac{1}{3}\) \(x_2 = \frac{{-5 - 7}}{6} = \frac{-12}{6} = -2\) The two solutions are \(x = \frac{1}{3}\) and \(x = -2\).
Quadratic Equation Solution
\[ \begin{align*} 3x^2 + 5x - 2 &= 0 \\ (3x - 1)(x + 2) &= 0 \\ \end{align*} \] \[ \begin{align*} 3x - 1 &= 0 \quad 或 \quad x + 2 = 0 \\ x &= \frac{1}{3} \quad 或 \quad x = -2 \\ \end{align*} \]
Solving Quadratic Equation x^2 + x - 20 = 0
Voy a resolver la ecuación de la parte a: x^2 + x - 20 = 0 Esta es una ecuación cuadrática que se puede factorizar para encontrar las soluciones. Buscamos dos números que multiplicados den -20 y sumados den +1 (el coeficiente del término lineal x). Estos números son +5 y -4, porque 5 * (-4) = -20 y 5 + (-4) = 1. Entonces podemos escribir la ecuación cuadrática como: (x + 5)(x - 4) = 0 Para encontrar las soluciones, igualamos cada factor a cero: x + 5 = 0 o x - 4 = 0 De la primera ecuación, x = -5. De la segunda ecuación, x = 4. Entonces las soluciones para la ecuación x^2 + x - 20 = 0 son x = -5 y x = 4.
Solving a Quadratic Equation by Factoring
The expression in the image is a quadratic equation: \[ 3n^2 + 9n - 30 = 0 \] To solve this quadratic equation, we can first attempt to factor it if it is factorable. If it's not easily factorable, we would use the quadratic formula. Let's see if the equation is factorable by looking for two numbers that multiply to \(3 \times (-30) = -90\) and add up to \(9\). These two numbers are 15 and -6 because \(15 \times -6 = -90\) and \(15 + (-6) = 9\). We can then express \(9n\) as \(15n - 6n\), rewriting the equation as: \[ 3n^2 + 15n - 6n - 30 = 0 \] Then we group the terms: \[ (3n^2 + 15n) - (6n + 30) = 0 \] Factoring by grouping, we take out the greatest common factor from each group: \[ 3n(n + 5) - 6(n + 5) = 0 \] Now, since both terms have a common factor of \((n + 5)\), we factor it out: \[ (n + 5)(3n - 6) = 0 \] Setting each factor equal to zero gives us our potential solutions: \[ n + 5 = 0 \quad \Rightarrow \quad n = -5 \] \[ 3n - 6 = 0 \quad \Rightarrow \quad 3n = 6 \quad \Rightarrow \quad n = 2 \] Therefore, the solutions to the quadratic equation are \(n = -5\) and \(n = 2\).
Finding the Value of a Coefficient in a Quadratic Equation
Bu denklemin köklerinin toplamı ve çarpımı, katsayıları kullanarak bulunabilir. Denklemin genel formu \( x^2 + bx + c = 0 \) olduğunda, köklerin toplamı \(-b\) ve köklerin çarpımı \(c\) olur. Denklemin verildiği haliyle, \(b = m - 1\) ve \(c = m + 1\)'dir. x_1 ve x_2 için verilen bilgilere göre: - Köklerin toplamı \( x_1 + x_2 = m - 1 \) olur ve bu \( 1 + 2 = 3 \) eşittir. - Köklerin çarpımı \( x_1 \cdot x_2 = m + 1 \) ve bu \(1 \cdot 2 = 2\)'dir. Bu yüzden iki denklemimiz şöyle olacak: 1. \( m - 1 = 3 \) 2. \( m + 1 = 2 \) Her iki denklemleri çözelim: 1. \( m - 1 = 3 \) \( m = 3 + 1 \) \( m = 4 \) 2. \( m + 1 = 2 \) \( m = 2 - 1 \) \( m = 1 \) Ancak \(m\) için iki farklı değer bulduk, ki bu mümkün değildir çünkü \(m\) tek bir değere sahip olmalıdır. Burada bir yanlışlık olmalı. Gerçekte, denklemimizden iki kök olduğuna ve köklerin çarpımının \(2\) olduğuna göre, ikinci denklemin doğru çözüm olduğunu anlarız. Buna göre, \(m\) için yalnızca \(1\) değeri mantıklıdır. Sonuç olarak: \( m = 1 \) Bu yüzden \(m\) nin alabileceği değer \(1\) ve sorulan \(m\) nin alabileceği değerler çarpımı da \(1\) dir.
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