Thrice the square of a natural number decreas | vedclass

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(B) Let the natural number be $x$.According to the problem,the equation is:$3x^2 - 4x = x + 50$Rearranging the terms to form a quadratic equation:$3x^

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$5$

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(B) Let the natural number be $x$.According to the problem,the equation is:$3x^2 - 4x = x + 50$Rearranging the terms to form a quadratic equation:$3x^2 - 5x - 50 = 0$Factoring the quadratic equation:$3x^2 - 15x + 10x - 50 = 0$$3x(x - 5) + 10(x - 5) = 0$$(3x + 10)(x - 5) = 0$This gives $x = -\frac{10}{3}$ or $x = 5$.Since the number must be a natural number,we discard the negative fraction.Therefore,the natural number is $5$.

Thrice the square of a natural number decreased by $4$ times the number is equal to $50$ more than the number. The number is

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