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Completing a square and quadratic equation solution
I Method - Methods of completing a square and to derive the formula for the solution of the quadratic equatio..
I Method - Methods of completing a square and to derive the formula for the solution of the quadratic equatio..Methods of solving quadratic equations
There are four methods of solving quadratic equations. i) By factorization ii) By completing the squares iii) By using the formula iv) By graphi..
II Method ( By Sridhar's Method)
Methods of completing a square and to derive the formula for the solution of the quadratic equation. ..
Methods of completing a square and to derive the formula for the solution of the quadratic equation. ..Type (ii) By expressing the polynomial as the difference of two squares
121x 2 - 25y 2 = (11x) 2 - (5y) 2 = (11x + 5y) (11x - 5y) [Using the identity a 2 -b 2 =(a-b)(a+b)] Factorise: (5a + 6b) 2 - 49b 2 Let x = 5a + 6b Then the given expression = (x) 2 - (7b) 2 = (x + 7b) (x - 7b) Re-substituting the value of x, we get = [(5a + 6b + 7b)] [(5a + 6b) - 7b]..
121x 2 - 25y 2 = (11x) 2 - (5y) 2 = (11x + 5y) (11x - 5y) [Using the identity a 2 -b 2 =(a-b)(a+b)] Factorise: (5a + 6b) 2 - 49b 2 Let x = 5a + 6b Then the given expression = (x) 2 - (7b) 2 = (x + 7b) (x - 7b) Re-substituting the value of x, we get = [(5a + 6b + 7b)] [(5a + 6b) - 7b]..Nature of the roots
Without solving the quadratic equation, the nature of the roots can be determined using the discriminant. i) D >0 i.e., positive and not a perfect square. The roots are real and distinct (irrational). ii) D >0 i.e., perfect square. The roots are rational and distin..
Type 5
Equations of the type This equation can be solved by squaring both the sides. This equation is solved by equating both sides, rearranging and squaring aga..
Equations of the type This equation can be solved by squaring both the sides. This equation is solved by equating both sides, rearranging and squaring aga..Nature of the roots
Without solving the quadratic equation, the nature of the roots can be determined using the discriminant. i) D >0 i.e., positive and not a perfect square. The roots are real and distinct (irrational). ii) D >0 i.e., perfect square. The roots are rational and distinct. iv) ..
Without solving the quadratic equation, the nature of the roots can be determined using the discriminant. i) D >0 i.e., positive and not a perfect square. The roots are real and distinct (irrational). ii) D >0 i.e., perfect square. The roots are rational and distinct. iv) ..Summary
If all the terms of the polynomial have a common factor, we take out the common factor and factorise. If the polynomial can be expressed as the difference of two squares, we use a 2 - b 2 = (a + b) (a - b)..
Suggested answer:
x+3y=6 x=6-3y y=2, x=6-3y =6-3(2) =6-6 x=0 y=0, x=6-3y =6-3(0) =6-0 x=6 y=+3, x=6-3y =6-3(3) =6-9 x=-3 a) Area between line and axes = area of D AOB =6 sq.units b) When x=-6, x = 6 - 3y , when x = -6, -6 = 6 - 3y -6 - 6 = - 3y -12 = - 3y \ if x = -6, y = 4..
x+3y=6 x=6-3y y=2, x=6-3y =6-3(2) =6-6 x=0 y=0, x=6-3y =6-3(0) =6-0 x=6 y=+3, x=6-3y =6-3(3) =6-9 x=-3 a) Area between line and axes = area of D AOB =6 sq.units b) When x=-6, x = 6 - 3y , when x = -6, -6 = 6 - 3y -6 - 6 = - 3y -12 = - 3y \ if x = -6, y = 4..Summary
If all the terms of the polynomial have a common factor, we take out the common factor and factorise . If all the terms of the polynomial have a common factor, we take out the common factor and factorise . If the polynomial can be expressed as the difference of two squares, we use ..
If all the terms of the polynomial have a common factor, we take out the common factor and factorise . If all the terms of the polynomial have a common factor, we take out the common factor and factorise . If the polynomial can be expressed as the difference of two squares, we use ..See what our Users say :
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