if x=2/3 and x=-3 are the roots of the QE ax^2+7x+b=0find value of A and b
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Given that x = 2/3 and x = -3 are the roots of the equation
=> x - 2/3 = 0 and x + 3 = 0 are the factors of the given polynomial ax² + 7x + b
Now,
x - 2/3 = 0
Multiply 3 on both sides
=> 3(x - 2/3) = 3(0)
=> 3x - 2 = 0
So 3x - 2 and x + 3 are the two factors of the given quadratic equation. Since a quadratic equation has only two factors, it means that when we will multiply the factors, the product will be the polynomial.
=> (3x - 2)(x + 3) = ax² + 7x + b
=> 3x² + 9x - 2x - 6 = ax² + 7x + b
=> 3x² + 7x - 6 = ax² + 7x + b
Now if we will compare them, then :-
In LHS, coefficient of x² is 3 and in RHS coefficient of x² is a
=> a = 3
Similarly, in LHS the constant term is - 6 and in RHs the constant term is b
=> b = -6
So your answer,
a = 3
b = - 6
Hope it helps dear friend ☺️✌️
=> x - 2/3 = 0 and x + 3 = 0 are the factors of the given polynomial ax² + 7x + b
Now,
x - 2/3 = 0
Multiply 3 on both sides
=> 3(x - 2/3) = 3(0)
=> 3x - 2 = 0
So 3x - 2 and x + 3 are the two factors of the given quadratic equation. Since a quadratic equation has only two factors, it means that when we will multiply the factors, the product will be the polynomial.
=> (3x - 2)(x + 3) = ax² + 7x + b
=> 3x² + 9x - 2x - 6 = ax² + 7x + b
=> 3x² + 7x - 6 = ax² + 7x + b
Now if we will compare them, then :-
In LHS, coefficient of x² is 3 and in RHS coefficient of x² is a
=> a = 3
Similarly, in LHS the constant term is - 6 and in RHs the constant term is b
=> b = -6
So your answer,
a = 3
b = - 6
Hope it helps dear friend ☺️✌️
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