Example Question - quadratic formula
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Solving Quadratic Equations
<p>Para resolver las ecuaciones cuadráticas se pueden utilizar distintos métodos como factorización, completar el cuadrado o la fórmula cuadrática. En este caso, aplicaremos la fórmula cuadrática \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \) para cada ecuación dado que no todos los términos son fácilmente factorizables.</p> <p>a) \( x^2 + 2x + 10 = 0 \)</p> <p>Usamos la fórmula cuadrática donde a = 1, b = 2, c = 10:</p> <p>\( x = \frac{{-2 \pm \sqrt{{2^2 - 4 \cdot 1 \cdot 10}}}}{{2 \cdot 1}} \)</p> <p>\( x = \frac{{-2 \pm \sqrt{{-36}}}}{{2}} \)</p> <p>\( x = \frac{{-2 \pm 6i}}{2} \)</p> <p>\( x = -1 \pm 3i \)</p> <p>b) \( x^2 + 4x + 29 = 0 \)</p> <p>Usamos la fórmula cuadrática donde a = 1, b = 4, c = 29:</p> <p>\( x = \frac{{-4 \pm \sqrt{{4^2 - 4 \cdot 1 \cdot 29}}}}{{2 \cdot 1}} \)</p> <p>\( x = \frac{{-4 \pm \sqrt{{-84}}}}{2} \)</p> <p>\( x = -2 \pm 2\sqrt{21}i \)</p> <p>c) \( x^2 - 6x + 13 = 0 \)</p> <p>Usamos la fórmula cuadrática donde a = 1, b = -6, c = 13:</p> <p>\( x = \frac{{6 \pm \sqrt{{(-6)^2 - 4 \cdot 1 \cdot 13}}}}{{2 \cdot 1}} \)</p> <p>\( x = \frac{{6 \pm \sqrt{{36 - 52}}}}{2} \)</p> <p>\( x = \frac{{6 \pm \sqrt{{-16}}}}{2} \)</p> <p>\( x = 3 \pm 2i \)</p> <p>d) \( 2x^2 + 12x + 68 = 0 \)</p> <p>Usamos la fórmula cuadrática donde a = 2, b = 12, c = 68:</p> <p>\( x = \frac{{-12 \pm \sqrt{{12^2 - 4 \cdot 2 \cdot 68}}}}{{2 \cdot 2}} \)</p> <p>\( x = \frac{{-12 \pm \sqrt{{144 - 544}}}}{4} \)</p> <p>\( x = \frac{{-12 \pm \sqrt{{-400}}}}{4} \)</p> <p>\( x = -3 \pm 5i \)</p> <p>e) \( 23 + 6x + x^2 = 0 \) (Reordenamos la ecuación para que tenga la forma estándar)</p> <p>\( x^2 + 6x + 23 = 0 \)</p> <p>Usamos la fórmula cuadrática donde a = 1, b = 6, c = 23:</p> <p>\( x = \frac{{-6 \pm \sqrt{{6^2 - 4 \cdot 1 \cdot 23}}}}{{2 \cdot 1}} \)</p> <p>\( x = \frac{{-6 \pm \sqrt{{36 - 92}}}}{2} \)</p> <p>\( x = \frac{{-6 \pm \sqrt{{-56}}}}{2} \)</p> <p>\( x = -3 \pm 2\sqrt{14}i \)</p>
Solving a Quadratic Equation
<p>We have the quadratic equation \(2x^2 - 4x - 6 = 0\).</p> <p>To solve for \(x\), we can use the quadratic formula: \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\), where \(a = 2\), \(b = -4\), and \(c = -6\).</p> <p>First, calculate the discriminant (\(b^2 - 4ac\)): \((-4)^2 - 4 \cdot 2 \cdot (-6) = 16 + 48 = 64\).</p> <p>Since the discriminant is positive, there are two real solutions.</p> <p>Now compute the two solutions using the quadratic formula:</p> <p>\(x = \frac{{-(-4) \pm \sqrt{64}}}{2 \cdot 2} = \frac{{4 \pm 8}}{4}\).</p> <p>So, the two solutions are:</p> <p>\(x = \frac{{4 + 8}}{4} = \frac{{12}}{4} = 3\)</p> <p>and</p> <p>\(x = \frac{{4 - 8}}{4} = \frac{{-4}}{4} = -1\).</p> <p>Thus, the solutions to the equation \(2x^2 - 4x - 6 = 0\) are \(x = 3\) and \(x = -1\).</p>
Quadratic Equation Solution Using the Quadratic Formula
The provided image shows the quadratic formula, which is used to find the solutions of a quadratic equation \( ax^2 + bx + c = 0 \). The solutions using the quadratic formula are: \[ x = \frac{{-b \pm \sqrt{b^2 - 4ac}}}{{2a}} \] This formula calculates the roots of any quadratic equation where \( a \neq 0 \). The term inside the square root, \( b^2 - 4ac \), is known as the discriminant, and it determines the nature of the roots: - If \( b^2 - 4ac > 0 \), there are two distinct real roots. - If \( b^2 - 4ac = 0 \), there is one real root (a repeated root). - If \( b^2 - 4ac < 0 \), there are two complex roots. To solve a specific quadratic equation using this formula, one should substitute the values of \( a \), \( b \), and \( c \) from the equation into the formula and simplify.
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 \(x = -2\).
Solving Quadratic Equation
确定 a=3，b=5 和 c=-2。 使用二次公式：x = [-b ± √(b²-4ac)] / (2a) 计算 b² - 4ac = 5² - 4×3×(-2) = 25 + 24 = 49 求根 √(b²-4ac) = √49 = 7 代入公式：x = [-5 ± 7] / (2×3) 解得两个解： x1 = (-5 + 7) / 6 = 2 / 6 = 1/3 x2 = (-5 - 7) / 6 = -12 / 6 = -2 所以，方程的解为 x = 1/3 或 x = -2。
Solving a Quadratic Equation by the Quadratic Formula
Para resolver la ecuación que aparece en la imagen, necesitamos simplificar y expandir las expresiones en ambos lados del signo igual, y después recolectar términos semejantes y resolver para x. La ecuación es: \( 6(2x - 3) + (2x - 9)^2 = (5x + 1)(4x - 3) \) Comenzamos expandiendo y simplificando ambos lados de la ecuación: Primero, el lado izquierdo de la ecuación: \( 6(2x - 3) = 12x - 18 \) \( (2x - 9)^2 = (2x - 9)(2x - 9) = 4x^2 - 18x - 18x + 81 = 4x^2 - 36x + 81 \) Sumamos ambas expresiones simplificadas: \( 12x - 18 + 4x^2 - 36x + 81 \) \( = 4x^2 - 24x + 63 \) Ahora, el lado derecho de la ecuación: \( (5x + 1)(4x - 3) = 20x^2 - 15x + 4x - 3 = 20x^2 - 11x - 3 \) Igualamos ambas expresiones: \( 4x^2 - 24x + 63 = 20x^2 - 11x - 3 \) Ahora, llevamos todos los términos al mismo lado para tener una ecuación cuadrática igualada a cero: \( 4x^2 - 24x + 63 - 20x^2 + 11x + 3 = 0 \) \( -16x^2 + 13x + 66 = 0 \) Esta es una ecuación cuadrática que podemos resolver por factorización, completando el cuadrado o usando la fórmula cuadrática. Sin embargo, al mirar los coeficientes, parece que no hay factores obvios, así que la fórmula cuadrática parece ser la mejor opción. La fórmula cuadrática es \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), donde \(a\), \(b\), y \(c\) son los coeficientes de la ecuación \(ax^2 + bx + c = 0\). En este caso, \(a = -16\), \(b = 13\), y \(c = 66\). Calculamos el discriminante (\(b^2 - 4ac\)): \(13^2 - 4(-16)(66) = 169 + 4224 = 4393\) Y aplicamos la fórmula cuadrática: \(x = \frac{-13 \pm \sqrt{4393}}{-32}\) Por lo tanto, tenemos dos soluciones posibles para \(x\), dependiendo de si tomamos la raíz cuadrada positiva o negativa del discriminante. Como el discriminante es un número positivo, sabemos que existen dos soluciones reales y distintas para esta ecuación cuadrática.
Solving Quadratic Equations with Coefficients
La ecuación que se muestra en la imagen es una ecuación cuadrática y se ve así: \[ 2x^2 - 4x + 1 = 0 \] Para resolver esta ecuación cuadrática, se puede utilizar la fórmula general para las raíces de una ecuación cuadrática, la cual es: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \] En donde \( a \), \( b \) y \( c \) son los coeficientes de la ecuación cuadrática. En este caso, tenemos que \( a = 2 \), \( b = -4 \), y \( c = 1 \). Primero calculamos el discriminante (\( \Delta \)), que es \( b^2 - 4ac \): \[ \Delta = (-4)^2 - 4(2)(1) = 16 - 8 = 8 \] Ahora usamos la fórmula general para encontrar los valores de \( x \): \[ x = \frac{{-(-4) \pm \sqrt{8}}}{{2(2)}} \] \[ x = \frac{{4 \pm \sqrt{8}}}{{4}} \] \[ x = \frac{{4 \pm 2\sqrt{2}}}{{4}} \] Podemos simplificar dividiendo numerador y denominador entre 2: \[ x = \frac{{2 \pm \sqrt{2}}}{{2}} \] De aquí obtenemos dos soluciones para la ecuación: \[ x_1 = \frac{{2 + \sqrt{2}}}{2} \] \[ x_2 = \frac{{2 - \sqrt{2}}}{2} \] Estas son las dos soluciones de la ecuación cuadrática dada.
Expanding a Quadratic Function
La función que se muestra en la imagen es \( f(x) = (2x^2 - 7x)^2 \). Para resolverla, tenemos que expandir la expresión. Esto lo hacemos aplicando la fórmula del cuadrado de un binomio, que es \( (a - b)^2 = a^2 - 2ab + b^2 \). Siguiendo esta fórmula y aplicándola a nuestra función, tendremos lo siguiente: \( f(x) = (2x^2 - 7x)^2 \) \( f(x) = (2x^2)^2 - 2 \cdot (2x^2)(7x) + (7x)^2 \) \( f(x) = 4x^4 - 28x^3 + 49x^2 \) Así que, al expandir la expresión original, obtenemos \( f(x) = 4x^4 - 28x^3 + 49x^2 \). Esto completa el cuadrado de la expresión dada.
Solving Quadratic Equation using Quadratic Formula
The equation provided in the image is a quadratic equation, which is in the form \(ax^2 + bx + c = 0\): \(3x^2 - 7x - 32 = 0\) To solve this quadratic equation, we can either factorise it, complete the square, or use the quadratic formula. The quadratic formula is: \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\) Here, \(a = 3\), \(b = -7\), and \(c = -32\). Let's use the quadratic formula: \(x = \frac{{-(-7) \pm \sqrt{{(-7)^2 - 4(3)(-32)}}}}{{2(3)}}\) \(x = \frac{{7 \pm \sqrt{{49 + 384}}}}{{6}}\) \(x = \frac{{7 \pm \sqrt{{433}}}}{{6}}\) Since \(433\) is a prime number, it cannot be simplified further into a perfect square, so the solutions to the equation are: \(x = \frac{{7 + \sqrt{433}}}{6}\) and \(x = \frac{{7 - \sqrt{433}}}{6}\) These are the roots of the quadratic equation \(3x^2 - 7x - 32 = 0\).
Quadratic Equation Solution
The given equation is a quadratic equation of the form ax^2 + bx + c = 0. To solve for x, we can use the quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / (2a) For our given equation: 3x^2 + 5x - 2 = 0 Here, a = 3, b = 5, and c = -2. Plugging these values into the quadratic formula gives us: x = [-5 ± sqrt(5^2 - 4 * 3 * (-2))] / (2 * 3) x = [-5 ± sqrt(25 + 24)] / 6 x = [-5 ± sqrt(49)] / 6 x = [-5 ± 7] / 6 This gives us two possible solutions for x: x = (-5 + 7) / 6 = 2 / 6 = 1/3 or x = (-5 - 7) / 6 = -12 / 6 = -2 Therefore, the solutions for x are x = 1/3 or x = -2.
Solving Quadratic Equations by Using the Quadratic Formula
The image you've provided is a little blurry, but it appears to show an algebraic expression that needs to be solved for x. The expression given is: 3x^2 + 5x - 2 = 0 This is a quadratic equation, which can be solved using several methods, such as factoring, completing the square, or using the quadratic formula. The quadratic formula states that for any quadratic equation in the form ax^2 + bx + c = 0, where a, b, and c are coefficients, the solutions for x can be found using: x = (-b ± √(b^2 - 4ac)) / (2a) In this case, a = 3, b = 5, and c = -2. Plugging these values into the formula gives us: x = (-5 ± √(5^2 - 4*3*(-2))) / (2*3) x = (-5 ± √(25 + 24)) / 6 x = (-5 ± √49) / 6 x = (-5 ± 7) / 6 This results in two possible solutions for x: x = (-5 + 7) / 6 x = 2 / 6 x = 1 / 3 or x = (-5 - 7) / 6 x = -12 / 6 x = -2 So the solutions for x are x = 1/3 or x = -2.
Solving a Quadratic Equation
Để giải phương trình thứ nhất: \[ x^4 - 12x^2 + 16 = 0 \] Chúng ta có thể đặt \( u = x^2 \), khi đó phương trình sẽ trở thành: \[ u^2 - 12u + 16 = 0 \] Đây là một phương trình bậc hai, ta có thể giải nó bằng cách sử dụng công thức nghiệm của phương trình bậc hai: \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Trong đó a = 1, b = -12, c = 16. Ta tính delta (\(\Delta\)): \[ \Delta = b^2 - 4ac = (-12)^2 - 4 \cdot 1 \cdot 16 = 144 - 64 = 80 \] Vì \( \Delta > 0 \), phương trình có hai nghiệm phân biệt: \[ u_1 = \frac{12 + \sqrt{80}}{2} = 6 + 2\sqrt{5} \] \[ u_2 = \frac{12 - \sqrt{80}}{2} = 6 - 2\sqrt{5} \] Bây giờ, vì \( u = x^2 \), ta cần lấy căn bậc hai của các nghiệm này để tìm nghiệm cho \( x \). Điều này cho ta bốn nghiệm: \[ x_1 = \sqrt{6 + 2\sqrt{5}} \] \[ x_2 = -\sqrt{6 + 2\sqrt{5}} \] \[ x_3 = \sqrt{6 - 2\sqrt{5}} \] \[ x_4 = -\sqrt{6 - 2\sqrt{5}} \] Đó là nghiệm của phương trình đầu tiên. Do đề bài yêu cầu giải bằng tiếng Việt, phần tiếng Anh chỉ để giải thích các bước đã làm.
Solving a Quadratic Equation Step by Step
Claro, resolvamos la ecuación paso a paso. La ecuación en la imagen es: \[ \frac{1}{x} + \frac{1}{2} (6x - 6) = -2 \] Primero, podemos simplificar la expresión eliminando el paréntesis distribuyendo \(\frac{1}{2}\) en los términos dentro del paréntesis: \[ \frac{1}{x} + \frac{1}{2} \cdot 6x - \frac{1}{2} \cdot 6 = -2 \] Esto se simplifica a: \[ \frac{1}{x} + 3x - 3 = -2 \] Ahora, queremos tener todos los términos en un lado de la ecuación y todo en términos de \(x\). Sumamos 3 a ambos lados de la ecuación: \[ \frac{1}{x} + 3x = 1 \] Para combinar los términos con \(x\), podemos multiplicar cada término por \(x\) para eliminar el denominador en el primer término: \[ 1 + 3x^2 = x \] Ahora, movemos todos los términos al lado izquierdo para obtener un polinomio igualado a cero: \[ 3x^2 - x + 1 = 0 \] Esta es una ecuación cuadrática y podemos resolverla usando la fórmula cuadrática: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Donde \(a = 3\), \(b = -1\) y \(c = 1\). Sustituimos los valores en la fórmula: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} \] \[ x = \frac{1 \pm \sqrt{1 - 12}}{6} \] \[ x = \frac{1 \pm \sqrt{-11}}{6} \] Dado que la raíz cuadrada de un número negativo no es un número real, la ecuación no tiene soluciones reales. Por lo tanto, \(x\) tendría dos soluciones complejas debido al discriminante negativo (\(-11\)). Estas soluciones están en la forma de números complejos.
Solving Quadratic Equations Using the Quadratic Formula
题目要求解方程 \(x^2 - 5x + 5 = 0\)。我们可以用求根公式（也称为二次方程的根的通用公式）来解这个方程。该公式为： \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] 其中，\(a\)、\(b\) 和 \(c\) 分别是方程 \(ax^2 + bx + c = 0\) 中的系数。在我们的例子中，\(a = 1\)、\(b = -5\)、\(c = 5\)。 根据这些值，我们可以将它们代入公式： \[x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}\] \[x = \frac{5 \pm \sqrt{25 - 20}}{2}\] \[x = \frac{5 \pm \sqrt{5}}{2}\] 所以，方程的解是： \[x_1 = \frac{5 + \sqrt{5}}{2}\] \[x_2 = \frac{5 - \sqrt{5}}{2}\] 这就是方程 \(x^2 - 5x + 5 = 0\) 的两个解。
Solving a Quadratic Equation by Factoring
The equation provided in the image is a quadratic equation that can be solved by factoring, completing the square, or using the quadratic formula. Since the coefficients are quite large, I'll solve it by factoring for simplicity. The equation is: \[ 28x^2 - 72x - 180 = 0 \] To make it easier to factor, we can look for a common factor between the coefficients, which in this case is 4. Let's divide the whole equation by 4 to simplify: \[ 7x^2 - 18x - 45 = 0 \] Now, we look to factor the quadratic. We need to find two numbers that multiply to \(7 \times (-45)\) (which is -315) and at the same time add up to -18 (the coefficient of the x term). These numbers are -25 and +7. So, we can express the middle term -18x as -25x + 7x, like this: \[ 7x^2 - 25x + 7x - 45 = 0 \] Group the terms to factor by grouping: \[ (7x^2 - 25x) + (7x - 45) = 0 \] Factor out the common factors: \[ x(7x - 25) + 5(7x - 25) = 0 \] Now you can see that \( (7x - 25) \) is a common factor: \[ (x + 5)(7x - 25) = 0 \] To find the solutions for x, set each factor equal to zero: \[ x + 5 = 0 \quad \text{or} \quad 7x - 25 = 0 \] Solve both equations: \[ x = -5 \quad \text{or} \quad 7x = 25 \] \[ \quad \text{or} \quad x = \frac{25}{7} \approx 3.571 \] So, the solutions to the quadratic equation are: \[ x = -5 \quad \text{or} \quad x = \frac{25}{7} \]
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