If 5 times of a positive whole number is less by 3 than twice of its square, then find the number?
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Let the number be x then 5x + 3 = 2x^2
This becomes a quadratic equation that is
2x^2 -5x - 3 = 0
2x^2 - 6x + x -3 =0
2x ( x-3 ) + 1 (x - 3 ) = 0
So (2x +1)(x-3) = 0
So by using common sense we can say that one of the two has to be zero then only can the product be zero so two cases now arise either
2x+1 = 0 wherein x=-1/2
OR
x-3=0 wherein x= 3
Now as the question states the value of x is a positive and a whole number then we can reject the value of -1/2 hence we get that the value of x is 3 . So the number being talked about in the question above is three .
This becomes a quadratic equation that is
2x^2 -5x - 3 = 0
2x^2 - 6x + x -3 =0
2x ( x-3 ) + 1 (x - 3 ) = 0
So (2x +1)(x-3) = 0
So by using common sense we can say that one of the two has to be zero then only can the product be zero so two cases now arise either
2x+1 = 0 wherein x=-1/2
OR
x-3=0 wherein x= 3
Now as the question states the value of x is a positive and a whole number then we can reject the value of -1/2 hence we get that the value of x is 3 . So the number being talked about in the question above is three .
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