Example Question - quadratic equation solutions
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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 Quadratic Equation with Square Root
The equation shown in the image appears to be a quadratic equation written as: \[ \sqrt{x^2 - 2x - 5} = x \] To solve this problem, we will first need to square both sides of the equation to eliminate the square root. Remember that when you square both sides of an equation, you may introduce extraneous solutions, so you'll need to check your answers in the original equation later. \[ (\sqrt{x^2 - 2x - 5})^2 = x^2 \] \[ x^2 - 2x - 5 = x^2 \] After squaring both sides, we notice that the x^2 terms on both sides of the equation are identical and they cancel out. \[ -2x - 5 = 0 \] Now, add 2x to both sides to isolate the constant term: \[ -5 = 2x \] To solve for x, divide both sides by 2: \[ x = -5 / 2 \] \[ x = -2.5 \] So the potential solution is x = -2.5. However, because we squared the equation to remove the square root, we need to verify that this potential solution satisfies the original equation: \[ \sqrt{(-2.5)^2 - 2(-2.5) - 5} \overset{?}{=} -2.5 \] Calculate the expression under the square root: \[ \sqrt{6.25 + 5 - 5} \] \[ \sqrt{6.25} \] We find the square root of 6.25: \[ \sqrt{6.25} = 2.5 \] Note that the square root of a positive number is positive, so the left side yields 2.5, not -2.5. Therefore, the potential solution x = -2.5 does not satisfy the original equation. This means that there is no solution to the original equation because the square root of a real number cannot equal a negative number. Thus, the original equation has no real solutions. If we needed to find complex solutions, we could continue the process knowing that the square root function would involve imaginary numbers when handling negative quantities, but based on the equation given, it seems we're dealing with real numbers only.
Solving a Quadratic Equation
Para resolver la ecuación cuadrática dada, la cual es \( x^2 + 8x - 5 \), podemos usar la fórmula general para ecuaciones cuadráticas, que es: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) Donde \( a \), \( b \) y \( c \) son los coeficientes de la ecuación cuadrática \( ax^2 + bx + c = 0 \). Para la ecuación \( x^2 + 8x - 5 \), los coeficientes serían: \( a = 1 \) \( b = 8 \) \( c = -5 \) Sustituimos estos valores en la fórmula general: \( x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} \) Calculamos dentro de la raíz cuadrada primero: \( 8^2 = 64 \) \( 4 \cdot 1 \cdot (-5) = -20 \) Lo cual nos da: \( x = \frac{-8 \pm \sqrt{64 - (-20)}}{2} \) \( x = \frac{-8 \pm \sqrt{64 + 20}}{2} \) \( x = \frac{-8 \pm \sqrt{84}}{2} \) La raíz cuadrada de 84 no es un número entero, así que podemos simplificarlo: \( \sqrt{84} = \sqrt{4 \cdot 21} = \sqrt{4} \cdot \sqrt{21} \) \( \sqrt{84} = 2 \cdot \sqrt{21} \) Entonces, la solución es: \( x = \frac{-8 \pm 2 \cdot \sqrt{21}}{2} \) Y simplificamos dividiendo por 2: \( x = -4 \pm \sqrt{21} \) Por lo tanto, las soluciones de la ecuación cuadrática son: \( x = -4 + \sqrt{21} \) y \( x = -4 - \sqrt{21} \)
Solving Quadratic Equations with the Quadratic Formula
To use the quadratic formula to solve the equation, we first need to put the equation in standard form, which is \( ax^2 + bx + c = 0 \). The equation given in the image is: \[ 4d^2 - 5d - 5 = 3d \] First, we will move all terms to one side to set the equation to zero: \[ 4d^2 - 5d - 5 - 3d = 0 \] \[ 4d^2 - 8d - 5 = 0 \] Now we have a quadratic equation in standard form, where \( a = 4 \), \( b = -8 \), and \( c = -5 \). The quadratic formula is: \[ d = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \] Let's plug the values of \( a \), \( b \), and \( c \) into the quadratic formula: \[ d = \frac{{-(-8) \pm \sqrt{{(-8)^2 - 4 \cdot 4 \cdot (-5)}}}}{2 \cdot 4} \] \[ d = \frac{{8 \pm \sqrt{{64 + 80}}}}{8} \] \[ d = \frac{{8 \pm \sqrt{{144}}}}{8} \] \[ d = \frac{{8 \pm 12}}{8} \] Now we have two potential solutions: \[ d = \frac{{8 + 12}}{8} = \frac{20}{8} = \frac{5}{2} \] \[ d = \frac{{8 - 12}}{8} = \frac{-4}{8} = \frac{-1}{2} \] Hence, the solutions to the given quadratic equation in simplest form are \( d = \frac{5}{2} \) and \( d = \frac{-1}{2} \).
Solving Quadratic Equations with Quadratic Formula
To solve the quadratic equation \( a(a - 3) = -1 \), you need to first expand the equation and then solve for \( a \). Step 1: Expand the left side of the equation: \( a \cdot a = a^2 \) \( a \cdot (-3) = -3a \) So, the expanded form is: \( a^2 - 3a \) Step 2: Bring all terms to one side to set the quadratic equation equal to zero: \( a^2 - 3a + 1 = 0 \) Step 3: Solve the quadratic equation. This can be done using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where in our equation, \( a = 1 \), \( b = -3 \), and \( c = 1 \). The quadratic formula gives: \( a = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \) \( a = \frac{3 \pm \sqrt{9 - 4}}{2} \) \( a = \frac{3 \pm \sqrt{5}}{2} \) So the solutions are: \( a = \frac{3 + \sqrt{5}}{2} \) or \( a = \frac{3 - \sqrt{5}}{2} \)
Solving Quadratic Equation with Square Root
The equation provided in the image is: √x - x = 0 To solve for x, start by isolating the square root term: √x = x Now, square both sides of the equation to eliminate the square root: (√x)^2 = x^2 This simplifies to: x = x^2 Rearrange the equation to set it to zero and find the values of x that satisfy the equation: x^2 - x = 0 Factor x out of the expression: x(x - 1) = 0 Now, set each factor equal to zero and solve for x: x = 0 x - 1 = 0 => x = 1 The solutions are x = 0 and x = 1. However, you must check these solutions in the original equation because squaring both sides can introduce extraneous solutions. For x = 0: √0 - 0 = 0 => 0 = 0 (true) For x = 1: √1 - 1 = 0 => 1 - 1 = 0 (true) Both solutions are valid, so x can be 0 or 1.
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