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How do you solve non-factorable equations? - Answers

If your equation is in the form ax2 + bx + c, it can always be solved for x by using the quadratic formula. x = -b plus or minus the square root of (b2 - 4ac) with the whole thing divided by 2a Let's try an easy example: x2 + 3x - 4 = 0 That's factorable: (x + 4)(x - 1) so we already know what the answers are (-4, 1) Now let's plug in the quadratic formula where a = 1, b = 3, c = -4 x = -3 plus or minus the square root of (3)2 -(4)1(-4) divided by 2(1) Simplified, that's -3 plus or minus the square root of 25 divided by 2 Our two answers are (-3 + 5) divided by 2 and (-3 - 5) divided by 2 2/2 = 1 and -8/2 = -4 Non-factorable equations work just as well, except the answer won't be a whole number. It'll be some nasty decimal or an imaginary number.



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How do you solve non-factorable equations? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_solve_non-factorable_equations

If your equation is in the form ax2 + bx + c, it can always be solved for x by using the quadratic formula. x = -b plus or minus the square root of (b2 - 4ac) with the whole thing divided by 2a Let's try an easy example: x2 + 3x - 4 = 0 That's factorable: (x + 4)(x - 1) so we already know what the answers are (-4, 1) Now let's plug in the quadratic formula where a = 1, b = 3, c = -4 x = -3 plus or minus the square root of (3)2 -(4)1(-4) divided by 2(1) Simplified, that's -3 plus or minus the square root of 25 divided by 2 Our two answers are (-3 + 5) divided by 2 and (-3 - 5) divided by 2 2/2 = 1 and -8/2 = -4 Non-factorable equations work just as well, except the answer won't be a whole number. It'll be some nasty decimal or an imaginary number.



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https://math.answers.com/math-and-arithmetic/How_do_you_solve_non-factorable_equations

How do you solve non-factorable equations? - Answers

If your equation is in the form ax2 + bx + c, it can always be solved for x by using the quadratic formula. x = -b plus or minus the square root of (b2 - 4ac) with the whole thing divided by 2a Let's try an easy example: x2 + 3x - 4 = 0 That's factorable: (x + 4)(x - 1) so we already know what the answers are (-4, 1) Now let's plug in the quadratic formula where a = 1, b = 3, c = -4 x = -3 plus or minus the square root of (3)2 -(4)1(-4) divided by 2(1) Simplified, that's -3 plus or minus the square root of 25 divided by 2 Our two answers are (-3 + 5) divided by 2 and (-3 - 5) divided by 2 2/2 = 1 and -8/2 = -4 Non-factorable equations work just as well, except the answer won't be a whole number. It'll be some nasty decimal or an imaginary number.

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      If your equation is in the form ax2 + bx + c, it can always be solved for x by using the quadratic formula. x = -b plus or minus the square root of (b2 - 4ac) with the whole thing divided by 2a Let's try an easy example: x2 + 3x - 4 = 0 That's factorable: (x + 4)(x - 1) so we already know what the answers are (-4, 1) Now let's plug in the quadratic formula where a = 1, b = 3, c = -4 x = -3 plus or minus the square root of (3)2 -(4)1(-4) divided by 2(1) Simplified, that's -3 plus or minus the square root of 25 divided by 2 Our two answers are (-3 + 5) divided by 2 and (-3 - 5) divided by 2 2/2 = 1 and -8/2 = -4 Non-factorable equations work just as well, except the answer won't be a whole number. It'll be some nasty decimal or an imaginary number.
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