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How do you solve x squared minus 4 y squared? - Answers
x^(2) - 4y^(2) Before calculating , we note #1 ' There is a negative sign between the two squares. So it will factor #2 ' '4' is the square of '2' . So we can redraw the terms as ( 2y)^2 Hence x^(2) - (2y)^2) This now satisfies the two conditions above. . So it will factor to ( x - 2y)(x + 2y) Note the different signs in each bracket . You hve asked for it to be solved!!! To solve you need to equate the terms to a value. In this case use '0' Hence (x - 2y)(x + 2y) = 0 Taking each bracket in turn x - 2y = 0 x = 2y or 2 = x/y or y = x/2 Then x + 2y = 0 x = -2y or -2/x/y or y = x/-2 Are the possible forms of solution. NB If you have two squared terms and a positive (+) between them, then it does NOT factor. NNB Think of the Pythagorean Equation a^2 + b^2 = h^2 This DOES NOT factor. However a^2 = h^2 - b^2 (Algebraic rearrangement) , DOES factor to a^2 = (h - b)(h + b)
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How do you solve x squared minus 4 y squared? - Answers
x^(2) - 4y^(2) Before calculating , we note #1 ' There is a negative sign between the two squares. So it will factor #2 ' '4' is the square of '2' . So we can redraw the terms as ( 2y)^2 Hence x^(2) - (2y)^2) This now satisfies the two conditions above. . So it will factor to ( x - 2y)(x + 2y) Note the different signs in each bracket . You hve asked for it to be solved!!! To solve you need to equate the terms to a value. In this case use '0' Hence (x - 2y)(x + 2y) = 0 Taking each bracket in turn x - 2y = 0 x = 2y or 2 = x/y or y = x/2 Then x + 2y = 0 x = -2y or -2/x/y or y = x/-2 Are the possible forms of solution. NB If you have two squared terms and a positive (+) between them, then it does NOT factor. NNB Think of the Pythagorean Equation a^2 + b^2 = h^2 This DOES NOT factor. However a^2 = h^2 - b^2 (Algebraic rearrangement) , DOES factor to a^2 = (h - b)(h + b)
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How do you solve x squared minus 4 y squared? - Answers
x^(2) - 4y^(2) Before calculating , we note #1 ' There is a negative sign between the two squares. So it will factor #2 ' '4' is the square of '2' . So we can redraw the terms as ( 2y)^2 Hence x^(2) - (2y)^2) This now satisfies the two conditions above. . So it will factor to ( x - 2y)(x + 2y) Note the different signs in each bracket . You hve asked for it to be solved!!! To solve you need to equate the terms to a value. In this case use '0' Hence (x - 2y)(x + 2y) = 0 Taking each bracket in turn x - 2y = 0 x = 2y or 2 = x/y or y = x/2 Then x + 2y = 0 x = -2y or -2/x/y or y = x/-2 Are the possible forms of solution. NB If you have two squared terms and a positive (+) between them, then it does NOT factor. NNB Think of the Pythagorean Equation a^2 + b^2 = h^2 This DOES NOT factor. However a^2 = h^2 - b^2 (Algebraic rearrangement) , DOES factor to a^2 = (h - b)(h + b)
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- og:descriptionx^(2) - 4y^(2) Before calculating , we note #1 ' There is a negative sign between the two squares. So it will factor #2 ' '4' is the square of '2' . So we can redraw the terms as ( 2y)^2 Hence x^(2) - (2y)^2) This now satisfies the two conditions above. . So it will factor to ( x - 2y)(x + 2y) Note the different signs in each bracket . You hve asked for it to be solved!!! To solve you need to equate the terms to a value. In this case use '0' Hence (x - 2y)(x + 2y) = 0 Taking each bracket in turn x - 2y = 0 x = 2y or 2 = x/y or y = x/2 Then x + 2y = 0 x = -2y or -2/x/y or y = x/-2 Are the possible forms of solution. NB If you have two squared terms and a positive (+) between them, then it does NOT factor. NNB Think of the Pythagorean Equation a^2 + b^2 = h^2 This DOES NOT factor. However a^2 = h^2 - b^2 (Algebraic rearrangement) , DOES factor to a^2 = (h - b)(h + b)
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