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How do you find the square root of 56? - Answers
There are several ways, but none (other than to use a calculator) are simple. Also, since sqrt(56) is irrational, you will not find an exact value for it.Bracket the result:Find two perfect squares on either side.49 < 56 < 64 so 7 < sqrt(56) < 8Next add two zeros:5476 < 5600 < 5625 ie 74 < sqrt(5600) < 75 so that 7.4 < sqrt(56) < 7.5Again559504 < 560000 < 561001 ie 748 < sqrt(560000) < 749 so that 7.48 < sqrt(56) < 7.49and again55995289 < 56000000 < 56010256 ie 7843 < sqrt(56000000) < 7844 so that 78.43 < sqrt(56) < 78.44You could continue.Newton-Raphson Method:You want the solution of x2 = 56equivalently, the zero for the equation f(x) = x2 - 56. Leaving aside the calculus for the rationale,You start with any guess for the answer, x0.Your next guess is x1 = x0 - (x02 - 56)/(2*x0)and the next one after that is calculated iteratively.If you started with x0 = 7the x1 = 7 - (72 - 56)/(2*7) = 7.5and so on, until very soon,x3 = 7.483314774, which is less than one in ten billionth off the correct answer.Finally, there is a method that resembles long division but I cannot explain it easily and I do not know what it is called so can't refer you to a site where it is described.
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How do you find the square root of 56? - Answers
There are several ways, but none (other than to use a calculator) are simple. Also, since sqrt(56) is irrational, you will not find an exact value for it.Bracket the result:Find two perfect squares on either side.49 < 56 < 64 so 7 < sqrt(56) < 8Next add two zeros:5476 < 5600 < 5625 ie 74 < sqrt(5600) < 75 so that 7.4 < sqrt(56) < 7.5Again559504 < 560000 < 561001 ie 748 < sqrt(560000) < 749 so that 7.48 < sqrt(56) < 7.49and again55995289 < 56000000 < 56010256 ie 7843 < sqrt(56000000) < 7844 so that 78.43 < sqrt(56) < 78.44You could continue.Newton-Raphson Method:You want the solution of x2 = 56equivalently, the zero for the equation f(x) = x2 - 56. Leaving aside the calculus for the rationale,You start with any guess for the answer, x0.Your next guess is x1 = x0 - (x02 - 56)/(2*x0)and the next one after that is calculated iteratively.If you started with x0 = 7the x1 = 7 - (72 - 56)/(2*7) = 7.5and so on, until very soon,x3 = 7.483314774, which is less than one in ten billionth off the correct answer.Finally, there is a method that resembles long division but I cannot explain it easily and I do not know what it is called so can't refer you to a site where it is described.
DuckDuckGo
How do you find the square root of 56? - Answers
There are several ways, but none (other than to use a calculator) are simple. Also, since sqrt(56) is irrational, you will not find an exact value for it.Bracket the result:Find two perfect squares on either side.49 < 56 < 64 so 7 < sqrt(56) < 8Next add two zeros:5476 < 5600 < 5625 ie 74 < sqrt(5600) < 75 so that 7.4 < sqrt(56) < 7.5Again559504 < 560000 < 561001 ie 748 < sqrt(560000) < 749 so that 7.48 < sqrt(56) < 7.49and again55995289 < 56000000 < 56010256 ie 7843 < sqrt(56000000) < 7844 so that 78.43 < sqrt(56) < 78.44You could continue.Newton-Raphson Method:You want the solution of x2 = 56equivalently, the zero for the equation f(x) = x2 - 56. Leaving aside the calculus for the rationale,You start with any guess for the answer, x0.Your next guess is x1 = x0 - (x02 - 56)/(2*x0)and the next one after that is calculated iteratively.If you started with x0 = 7the x1 = 7 - (72 - 56)/(2*7) = 7.5and so on, until very soon,x3 = 7.483314774, which is less than one in ten billionth off the correct answer.Finally, there is a method that resembles long division but I cannot explain it easily and I do not know what it is called so can't refer you to a site where it is described.
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- og:descriptionThere are several ways, but none (other than to use a calculator) are simple. Also, since sqrt(56) is irrational, you will not find an exact value for it.Bracket the result:Find two perfect squares on either side.49 < 56 < 64 so 7 < sqrt(56) < 8Next add two zeros:5476 < 5600 < 5625 ie 74 < sqrt(5600) < 75 so that 7.4 < sqrt(56) < 7.5Again559504 < 560000 < 561001 ie 748 < sqrt(560000) < 749 so that 7.48 < sqrt(56) < 7.49and again55995289 < 56000000 < 56010256 ie 7843 < sqrt(56000000) < 7844 so that 78.43 < sqrt(56) < 78.44You could continue.Newton-Raphson Method:You want the solution of x2 = 56equivalently, the zero for the equation f(x) = x2 - 56. Leaving aside the calculus for the rationale,You start with any guess for the answer, x0.Your next guess is x1 = x0 - (x02 - 56)/(2*x0)and the next one after that is calculated iteratively.If you started with x0 = 7the x1 = 7 - (72 - 56)/(2*7) = 7.5and so on, until very soon,x3 = 7.483314774, which is less than one in ten billionth off the correct answer.Finally, there is a method that resembles long division but I cannot explain it easily and I do not know what it is called so can't refer you to a site where it is described.
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