Example Question - completing the square
Here are examples of questions we've helped users solve.
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>
Converting Circle Equation to Center-Radius Form
The equation written in the image appears to be: \[ x^2 + y^2 + 2x - 4y - 11 = 0 \] This looks to be the equation of a circle in the general form, which can be converted into the center-radius form by completing the square for both x and y. The general equation for a circle is: \[ (x - h)^2 + (y - k)^2 = r^2 \] Where (h,k) is the center of the circle and r is its radius. Let's complete the square for the given equation: 1. Group x and y terms together and add and subtract the necessary constants inside the parenthesis to form a perfect square trinomial: \[ (x^2 + 2x + 1) + (y^2 - 4y + 4) - 1 - 4 - 11 = 0 \] 2. Complete the square by adding and subtracting the right constants inside each parenthesis: \[ (x^2 + 2x + 1) + (y^2 - 4y + 4) = 1 + 4 + 11 \] 3. Rewrite as squares and simplify the right side of the equation: \[ (x + 1)^2 + (y - 2)^2 = 16 \] Thus, the center-radius form of the given circle's equation is: \[ (x + 1)^2 + (y - 2)^2 = 4^2 \] So the center of the circle is (-1, 2) and the radius is 4.
Solving Quadratic Equations by Factoring
The equation provided in the image is a quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a = 6 \), \( b = -7 \), and \( c = -10 \). To solve this equation, we can either factor the quadratic, complete the square, or use the quadratic formula. Let's attempt factoring first. We need two numbers that multiply to \( a \times c = 6 \times (-10) = -60 \) and that add to \( b = -7 \). Looking for such numbers, we find that -12 and +5 multiply to -60 and add to -7. Now we can write: \(6x^2 - 12x + 5x - 10 = 0\) Group the terms to factor by grouping: \( (6x^2 - 12x) + (5x - 10) = 0 \) Factor out common factors: \( 6x(x - 2) + 5(x - 2) = 0 \) Now we can see that \( (x - 2) \) is a common factor: \( (6x + 5)(x - 2) = 0 \) Using the zero product property, we can set each factor equal to zero: \( 6x + 5 = 0 \) or \( x - 2 = 0 \) Solving for \( x \) gives us: \( x = -\frac{5}{6} \) or \( x = 2 \) So the solutions to the quadratic equation are \( x = -\frac{5}{6} \) and \( x = 2 \).
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