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How do you find the roots 3x2-5x equals 2? - Answers

First, we'll rearrange the expression in the standard format of ax2 + bx + c = 0: 3x2 - 5x = 2 3x2 - 5x - 2 = 0 Now we need to find two numbers whose sum is negative five (coefficient of the second term), and whose product is negative 6 (product of the coefficients of the first and last terms). Those two numbers in this case would be negative six and one. We'll break the middle term down into two terms using those coefficients: 3x2 - 6x + x - 2 = 0 Now we factor out a common multiple between our first two and last two terms: 3x(x - 2) + 1(x - 2) = 0 And we can group common terms: (3x + 1)(x - 2) = 0 Giving us the fully factored expression on the left.



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How do you find the roots 3x2-5x equals 2? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_find_the_roots_3x2-5x_equals_2

First, we'll rearrange the expression in the standard format of ax2 + bx + c = 0: 3x2 - 5x = 2 3x2 - 5x - 2 = 0 Now we need to find two numbers whose sum is negative five (coefficient of the second term), and whose product is negative 6 (product of the coefficients of the first and last terms). Those two numbers in this case would be negative six and one. We'll break the middle term down into two terms using those coefficients: 3x2 - 6x + x - 2 = 0 Now we factor out a common multiple between our first two and last two terms: 3x(x - 2) + 1(x - 2) = 0 And we can group common terms: (3x + 1)(x - 2) = 0 Giving us the fully factored expression on the left.



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https://math.answers.com/math-and-arithmetic/How_do_you_find_the_roots_3x2-5x_equals_2

How do you find the roots 3x2-5x equals 2? - Answers

First, we'll rearrange the expression in the standard format of ax2 + bx + c = 0: 3x2 - 5x = 2 3x2 - 5x - 2 = 0 Now we need to find two numbers whose sum is negative five (coefficient of the second term), and whose product is negative 6 (product of the coefficients of the first and last terms). Those two numbers in this case would be negative six and one. We'll break the middle term down into two terms using those coefficients: 3x2 - 6x + x - 2 = 0 Now we factor out a common multiple between our first two and last two terms: 3x(x - 2) + 1(x - 2) = 0 And we can group common terms: (3x + 1)(x - 2) = 0 Giving us the fully factored expression on the left.

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      First, we'll rearrange the expression in the standard format of ax2 + bx + c = 0: 3x2 - 5x = 2 3x2 - 5x - 2 = 0 Now we need to find two numbers whose sum is negative five (coefficient of the second term), and whose product is negative 6 (product of the coefficients of the first and last terms). Those two numbers in this case would be negative six and one. We'll break the middle term down into two terms using those coefficients: 3x2 - 6x + x - 2 = 0 Now we factor out a common multiple between our first two and last two terms: 3x(x - 2) + 1(x - 2) = 0 And we can group common terms: (3x + 1)(x - 2) = 0 Giving us the fully factored expression on the left.
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