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MATHS POWER POINT PRESENTATION MADE BY : SAHITYA GUPTA TOPIC : QUADRATIC EQUATION CLASS : 10th 'B'Slide 2:
I WILL TAKE YOU THE WORLD OF QUADRATIC EQUATION COME WITH MEQuadratic Equations:
Quadratic Equations An equation in the form a x 2 + b x + c = 0 where a, b, and c are real numbers and a≠0 is called a quadratic or a second degree equation.Slide 5:
Consider the quadratic equation 2 x 2 – 3 x + 1 = 0. If we replace x by 1 on the LHS of this equation, we get (2 × 12) – (3 × 1) + 1 = 0 = RHS of the equation. We say that 1 is a root of the quadratic equation 2 x 2 – 3 x + 1 = 0. This also means that 1 is a zero of the quadratic polynomial 2 x 2 – 3 x + 1. In general, a real number α is called a root of the quadratic equation ax 2 + bx + c = 0, a ≠ 0 if a α2 + b α + c = 0. We also say that x = α is a solution of the quadratic equation , or that α satisfies the quadratic equation . Note that the zeroes of the quadratic polynomial ax 2 + bx + c and the roots of the quadratic equation ax 2 + bx + c = 0 are the same . 1. Solve (x – 3)(x – 4) = 0. :
1. Solve ( x – 3)( x – 4) = 0. x = 3, 4 Solution: x – 3 = 0 or x – 4 = 0 x = 3 or x = 4 2. Solve x2 + 5x + 6 = 0.:
2. Solve x 2 + 5 x + 6 = 0. The solution to x 2 + 5 x + 6 = 0 is x = –3, –2 Solution: x 2 + 5 x + 6 = ( x + 2)( x + 3) Set this equal to zero: ( x + 2)( x + 3) = 0 Solve each factor: x + 2 = 0 or x + 3 = 0 x = –2 or x = – 33. Solve x2 – 3 = 2x. :
3. Solve x 2 – 3 = 2 x . The solution to x 2 – 3 = 2 x is x = –1, 3 get all the terms over on one side, with zero on the other side, then factor and solve: x 2 – 3 = 2 x x 2 – 2 x – 3 = 0 ( x – 3)( x + 1) = 0 x – 3 = 0 or x + 1 = 0 x = 3 or x = –14. Solve (x + 2)(x + 3) = 12. :
4. Solve ( x + 2)( x + 3) = 12. Multiply and simplify the left-hand side ( x + 2)( x + 3) = 12 x 2 + 5 x + 6 = 12 Subtract the 12 over to the left-hand side x 2 + 5 x + 6 – 12 = 0 x 2 + 5 x – 6 = 0 Re-factor, then solve. ( x + 6)( x – 1) = 0x + 6 = 0 or x – 1 = 0 x = –6 or x = 1 :
x + 6 = 0 or x – 1 = 0 x = –6 or x = 1 The solution to ( x + 2)( x + 3) = 12 is x = –6, 15. Solve x(x + 5) = 0:
5. Solve x ( x + 5) = 0 x = 0 or x + 5 = 0 x = 0 or x = –5 Solution: x ( x + 5) = 0 The solution to x ( x + 5) = 0 is x = 0, –56. Solve x2 – 5x = 0:
6. Solve x 2 – 5 x = 0 Then the solution to x 2 – 5 x = 0 is x = 0, 5 x ( x – 5) = 0 x = 0 or x – 5 = 0 x = 0 or x = 57. Solve x2 – 4 = 0. :
7. Solve x 2 – 4 = 0. Then the solution is x = –2, 2 or x = ± 2 x 2 – 4 = 0 ( x – 2)( x + 2) = 0 x – 2 = 0 or x + 2 = 0 x = 2 or x = –21. x2 – 4 = 0:
1. x 2 – 4 = 0 x = ± 2 Then the solution is x = ± 2 x 2 = 42. Solve x2 – 50 = 0:
2. Solve x 2 – 50 = 0 x 2 = 50 Then the solution is x 2 – 50 = 03. Solve (x – 5)2 – 100 = 0. :
3. Solve ( x – 5) 2 – 100 = 0. x = 5 ± 10 ( x – 5) 2 – 100 = 0 ( x – 5) 2 = 100 x – 5 = ±10x = 5 – 10 or x = 5 + 10 x = –5 or x = 15 :
x = 5 – 10 or x = 5 + 10 x = –5 or x = 15 The solution is x = –5, 154. Solve (x – 2)2 – 12 = 0 :
4. Solve ( x – 2) 2 – 12 = 0 ( x – 2) 2 – 12 = 0 ( x – 2) 2 = 12 Then the solution isSlide 21:
The Quadratic Formula: For ax 2 + bx + c = 0, the value of x is given bySlide 22:
Use the Quadratic Formula to solve 1.x 2 – 4 x – 8 = 0. a = 1, b = –4, and c = –8. 2. x 2 – 2 x – 24 = 0 3. x 2 – 2 x – 2 = 0 4. x 2 + 6 x +3 = 0 5. x 2 +3 x – 2 = 0 6. 3 x 2 + 4 x = 4EXAMPLE QUESTION:
EXAMPLE QUESTIONSolve the following quadratic equations.:
Solve the following quadratic equations. x 2 =5 x 16x 2 + 12 x = 0 x 2 - 36 = 0 81x 2 – 100= 0Solve the following quadratic equations by factoring.:
Solve the following quadratic equations by factoring. x 2 + x = 12 6x 2 = 11 x + 10Slide 26:
8x 2 – 10 x + 3 = 0 x 2 + 7 = 4 x Solve the following quadratic equations by quadratic formula.Slide 27:
I THINK SO YOU HAVE ENJOYED MY PRESENTATION THANK YOU BYE