Quadraticequation
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Quadraticequation

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Quadraticequation Quadraticequation Presentation Transcript

  • A QUADRATIC is a polynomialwhose highestexponent is 2.
  • A quadratic equation is a second-orderpolynomial equation in a single variable x ax2+bx+c=0with a ≠ 0. Because it is a second-orderpolynomial equation, the fundamentaltheorem of algebra guarantees that ithas two solutions. These solutions may be both real, or both complex.
  • The roots can be found by completing the square, Solving for then givesThis equation is known as the quadratic formula.
  • The plus-minus sign states that you have two numbers and
  • What do we mean by a root of a quadratic? A solution to the quadratic equation. For example, the roots of this quadratic x² + 2x − 8 are the solutions to x² + 2x − 8 = 0.
  • To find the roots, we can factor that quadratic as (x + 4)(x − 2).Now, if x = −4, then the first factorwill be 0. While if x = 2, the secondfactor will be 0. But if any factor is 0, then the entire product will be 0. Therefore, if x = −4 or 2, then x² + 2x − 8 = 0.
  • −4 and 2are the solutions to the quadraticequation. They are the roots of that quadratic.
  • A root of a quadraticis also called a zero. Because, as we will see, at each root, the value of the graph is 0.
  • How many roots has a quadratic?Always two. Because a quadratic (with leading coefficient 1, at least) can always be factored as (x − a)(x − b), and a, b are the two roots.Note that if a factor is (x + q), then the root is −q. For, (x + q) can take the form (x − a): (x + q) = [x − (−q)]. −q is the root,
  • What do we mean by a double root? The two roots are equal. The factors are(x − a)(x − a), so that the two roots are a, a. For example, this quadratic x² − 10x + 25 can be factored as (x − 5)(x − 5). If x = 5, then each factor will be 0, andtherefore the quadratic will be 0. 5 is called a double root.
  • When will a quadratic have a double root? When the quadratic is a perfect square trinomial.
  • Find the roots of eachquadratic by factoring.
  • a) x² − 3x + 2 (x − 1)(x − 2) x = 1 or 2.b) x² + 7x + 12 (x + 3)(x + 4) x = −3 or −4.
  • c) x² + 3x − 10 (x + 5)(x − 2)x = −5 or 2.d) x² − x – 30 (x + 5)(x − 6)x = −5 or 6.
  • e) 2x² + 7x + 3 (2x + 1)(x + 3)x = − 1 or −3. 2 f) 3x² + x − 2 (3x − 2)(x + 1)x= 2 or −1. 3
  • g) x² + 12x + 36 (x + 6)² x = −6, −6.A double root. h) x² − 2x + 1 (x − 1)² x = 1, 1.A double root.
  • Notice that we usethe conjunction "or," because x takes on only one value at a time.
  • c = 0. Solve thisquadratic equation: ax² + bx = 0
  • Since there is no constant term -- c = 0 -- x is a common factor: x(ax + b) = 0. This implies: x= 0 or x = b a. Those are the two roots.
  • a) x² − 5x x(x − 5)x = 0 or 5. b) x² + x x(x + 1)x = 0 or −1.
  • c) 3x² + 4x x(3x + 4)x = 0 or −4 3 d) 2x² − x x(2x − 1)x = 0 or ½
  • b = 0. Solve thisquadratic equation: ax² − c = 0.
  • In the case where there is no middle term, we can write: ax²=c. This implies: x²=c a x=
  • However, if the form is the difference of two squares -- x² − 16 -- then we can factor it as: (x + 4)(x −4). The roots are ±4. In fact, if the quadratic is x² − c, then we could factor it as: (x + )(x − ), so that the roots are ± .
  • a) x² − 3 x² = 3 x=± .b) x² − 25(x + 5)(x − 5) x = ±5. c) x² − 10(x + )(x − ) x=± .
  • Solve each equation for x.
  • a) x² = 5x − 6 x² − 5x + 6 = 0(x − 2)(x − 3) = 0 x = 2 or 3.b) x² + 12 = 8xx² − 8x + 12 = 0(x − 2)(x − 6) = 0 x = 2 or 6.
  • c) 3x² + x = 10 3x² + x − 10 = 0 (3x − 5)(x + 2) = 0 x = 5/3 or − 2. d) 2x² = x 2x² − x = 0 x(2x − 1) = 0 x = 0 or 1/2.
  • Solve this equation
  • We can put this equation in thestandard form by changing all the signs on both sides. 0 will not change. We have the standard form: 5 3− x − 3x² = 0 2
  • Next, we can get rid of the fraction by multiplying both sides by 2. Again, 0 will not change. 6x² + 5x − 6 = 0 (3x − 2)(2x + 3) = 0. 2 3 The roots are 3 and − 2
  • THE END.