How to Solve a Quadratic Equation by Graphing
2 – 6x + 9 = 0 by graphing Graph the related function f(x) = x – 6x + 9 x f(x) 0 9 1 4 2 1 3 0 4 1 5 4 6 9 Notice that the vertex of the parabola is the x-intercept Thus, one solution is 3 What is the other solution? Try solving the equation by factoring x – 6x + 9 = 0 Original equation (x – 3)(x – 3) = 0 Factor
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Solve x2 — 6x + 9 = 0 by graphing Graph the related function fix) = x The equation of the axis of symmetry is or 3 The vertex is at (3, 0) Graph 2(1) the vertex and several other points on either side of the axis of symmetry To solve x2 — 6x + 9 = 0, you need to know where the value of fix) 0 The vertex of the parabola is the x-intercept
Roots of Quadratic Equations - RIT
x 2 6x 9 0 a 1 b 6 c 9 a b b ac x 2 r 2 4 2(1) 6 r ( 6)2 4(1)(9) x 2 6r 36 36 x 2 6r 0 x 2 6r 0 x 3 2 6 x (There is only 1 real root ) b x 2 3x 1 0 b 3 c 1 2(1) 3r 9 4(1)(1) x 2 3r 5 x 2 3 5 x and 2 3 5 x The Discriminant Equals Zero The Discriminant is positive Since the discriminant is positive (it equals +5) there are two real roots
Quadratic and Exponential Functions
• Use the Quadratic Formula to solve –x2 + 6x – 9 = 0 Example • Use the Quadratic Formula to solve -3x2 + 6x + 9 = 0 Example
H 33 Finding Complex Solutions of Quadratic Equationsnotebook
The graph off(x) — x2 + 6x is shown Use the graph to determine how many real solutions the following equations have: x2 + 6x + 6 — 0, x2 + 6x 9 — 0, and x2 + 6x + 12 — 0 Explain Personal èãth TrSiner Online Homework Hints and Help Extra Practice 2 The graph of f(x) = —Lx2 + 3x is shown Use the graph to determine how
MAXIMUM AND MINIMUM 2 POINT OF INFLECTION
x3 6x2 9x = 0 i e When x x2 6x 9 = 0 (did you spot the factor x? It is one of the points you have to watch out for) So dy dx = 0 when x = 0 or when x2 6 9 = 0 Now x2 6x 9 = x 3 2 which is zero when dx x = 3 (or use the formula) So dy = 4x x 3 Using the formula for y, when x = 0 y = -7, and when x = 3 y = -7, so the stationary points are (0
CLASS-X MATHEMATICS WORKSHEET CHAPTER-4: QUADRATIC EQUATIONS
Q1 Show that x = -3 is the solution of the equation x2 +6x +9 = 0 Q2 For what value of k are the roots of quadratic equation 3x2 +2kx +27 = 0 real and equal? Q3 Write the nature of roots of quadratic equation 4x2 +4√3x +3 = 0 Q4 If a and b are the roots of the equation x2 +ax –b = 0, then find a and b Q5
2 QUADRATIC EQUATIONS
(x + 2 )2 – 9 = 0 ( x + 2)2 = 9 x + 2 = 3 x = -2 3 x = -5 or x = 1 L1 Solve x2 + 4x + 3 = 0 by method of ‘completing the square’ (Ans : – 1 , – 3 ) C2 Solve x2 – 6x + 3 = 0 by method of ‘completing the square’ Give your answer correct to 3 decimal places x2 – 6x + 3 = 0 3 2 6 2 6 6 2 2 2 x x = 0
FACTORIZACIÓN Y FRACCIONES ALGEBRAICAS RESUELTAS
6x +9 — (x 3)2, ya que las raíces de x2 6x-+9 0 son 0 son: 6+ 36-36 6 3 Raíz doble 3), ya que las raíces de +2x— 15 10 2+8 (F -25 Author: Hermanas Sardina
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