Solving Quadratic Equations - Speech or Presentation Example

Math Problem A quadratic equation is a polynomial expression with one variable. Finding solution of a quadratic equation denotes determination of the value of the variable for which the left side is equal to the right side. Solve equation:(y – 2/3 ) 2 = 4/9 . Solution of this equation will be achieved thru factoring. The concept factoring indicates multiplication of two binomial expressions (“Factoring quadratics”). Step-by-step solution is shown below.Step 1: Expand the left side by completing the square: (y-2/3) 2 = y 2 – (2* y*2/3) + (2/3*2/3) = y 2 - 4/3 y + 4 / 9Step 2: Rewrite the main equation: y 2 – 4/3 y + 4/9 = 4/9Step 3: Subtract 4/9 from both sides: y 2 – 4/3 y + (4/9 – 4/9) = (4/9 – 4/9) = y 2 – 4/3 y = 0.Step 4. Multiply both sides of the equation by 3: 3 * (y 2 – 4/3 y) = 3 * 0 , or 3 y 2 – 4 y = 0.

This is a form of quadratic equation: ax 2 – bx + c, where: a = 3, b = 4, and c = 0. The next steps show factorization of this equation. Step 5: Factorization of the coefficient “a” : 3 = 1 * 3Step 6: Factorization of the coefficient “c” : 0 = 0 * 4. Note: The selection is made so “b” can be 4.Step 7 : Arranging binomials: (1 0 ) * (3 4) Note: sign is not considered in this step.Step 8: Complete binomials with the sign: (y + 0) * (3 y – 4) = y (3y –4)Sep 9: Insert expression of step 8 in step 4: y ( 3y – 4) = 0Step 10: Using zero factor property: y = 0 or 3y – 4 = 0Step 11: Rewriting 3y – 4: Add 4 on both sides of 3y-4=0.

This gives, 3y=4. Divide both sides by 3. This gives , y = 4 /3 Step 11: Solutions: y = 0 or y = 4/3Step 12: Solution set {0 , 4/3}Check, (y – 2/3) 2 = 4/9Step 1. For the value, y = 0. Substitute zero for y, we get: (0 – 2/3) 2 = 4/9, or 4/9 = 4/9Sep 2. For the value y = 4/3. Substitute 4 /3 for y, we get: (4/3 – 2/3) 2 = 4/9, or (2/3) 2 = 4/9, or 4/9 = 4/9. The quadratic formula helps to find the solution of a quadratic equation. A quadratic equation is expressed as; ax 2 + bx + c = 0.

Using quadratic formula, solutions can be derived in the form, -b + / √ b 2 - 4 ac X = ---------------------------- . 2aIn this equation b2 – 4ac is called discriminant. It is determined using three coefficients of the equation. The value of “a” cannot be zero. The value of discriminant can be less than zero, greater than zero, equal to zero, and positive perfect square. Solve equation , 2 x 2 + 5 x – 1 = 0 Step 1: Coefficients : a = 2, b = 5, and c= - 1.

Solution will be achieved using formula, - (5) + / √ [ (5) 2 - (4 * 2 * -1)] X = ------------------------------------------ 2 (2) Step 2. Discriminant , b 2 – 4ac : (5) 2 - 4 * 2 * -1 = 25 + 8 = 33. Step 3. Find square root: √ 33 = 5.744 Step 4 . First solution: (- 5 + 5.744) / 4 = 0.186Step 5. Second solution: (- 5 – 5.744) / 4 = - 2.686Step 6 . Solution set. { 0.186 , - 2.686) ReferenceFactoring quadratics. (n.d). mathisfun. Retrieved from http://www.

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