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# Finding the roots of an equation? Topic: Problem solving by quadratic equations
July 16, 2019 / By Arran
Question: How do you find the roots of the equation 3x^2 = -2x - 3. Gotten somewhat into the problem but can't seem to completely solve it. Any help would be greatly appreciated. How come I got up to x = -2+- sq(4 - (-36) / 6 ## Best Answers: Finding the roots of an equation? Unity | 8 days ago
The roots are the solution to the equation. 3x^2 = -2x - 3 3x^2 + 2x + 3 = 0 This quadratic equation cannot be solve by factoring. This means you must use the quadratric formula. To learn about the quadratic formula, see link: http://www.purplemath.com/modules/quadfo... ===================================== After doing the math, I found the roots to be: Final answers: x = [-1 + 2(sqrt{2}) i]/3 AND x = [-1 - 2(sqrt{2}) i]/3 NOTE: The roots of your equation are complex roots.
👍 284 | 👎 8
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We found more questions related to the topic: Problem solving by quadratic equations Originally Answered: How do you write an equation of a line? please help? Sarina
this is a quardatic eqn. you can id'fy one, if the highest power of x is 2. so, bring all the terms to the left hand side, by which the eqn becomes 3x^2+2x+3=0. now, compare with the general form: ax^2+bx+c=0, and you will see a is 3, b is 2, and c is 3. to find x that will have two values which CAN be identical, solve this problem: the numerator is -b+ square root of [b^2-4ac], and the denominator is 2a. also do the problem where the numerator is -b - sq. root of [b^2-4ac].the denominator is 2a. both the problems will give the values of x. as said earlier, the answers can be identical. remember, if b^2-4ac is a negative number, just find the root of the number ignoring the sign, and write "i" at the end. eg. sq.root of -4 is 2i. hope this helped.
👍 120 | 👎 0 Norah
Given equation is x^2 - 11x + 28 = 0 Now evaluate the climate of the consistent fee such that whilst they're further or subtracted we get 11 through fact the cost so 7 * 4 = 28 and seven + 4 = 11 may well be seen be careful specifically circumstances we could get 2 or greater pairs like the comparable element then we could use all of them and look at which one ends up in answer now x^2 -4x -7x +28=0 (here split 11 in terms of the climate) take x as hardship-loose from 1st area and seven as hardship-loose from 2d area x(x-4)-7(x-4)=0 here we could see that (x-4) could be contemporary in the two areas (x-7)(x-4)=0 now equate the two equations seperately to 0 x-7=0 x-4=0 x=7 x=4 so 7 and four are the roots of quadratic equation the above defined technique is one technique we've one technique that's that if the equation is in variety ax^2 + bx + c = 0 Then the roots may well be found out utilising the formulation x = (-b + (b^2 -4ac)^a million/2 )/2a or (-b - (b^2 -4ac)^a million/2 )/2a so here in given equation we've a=a million b=-11 c=28 now if we persist with in the above formulation x=(11+(11^2-(28*4*a million))^a million/2))/2*a million =(11+(121-112)^a million/2))/2 =(11+(9)^a million/2)/2 =(11+3)/2 =7 x=(11-(11^2-(28*4*a million))^a million/2))/2*a million =(11-(121-112)^a million/2))/2 =(11-(9)^a million/2)/2 =(11-3)/2 =4 so roots are 7 and four now did u get it! bye
👍 113 | 👎 -8 Mabella
Solve for zero: 3x² + 2x + 3 = 0 This doesn't factor so use the quadratic formula or completing the square: Quadratic Formula: (-2 ± √(2² - 4(3)(3))) / 2(3) = (-2 ± 4i√2) / 6 = (-1 ± 2i√2) / 3 Completing the square: 3x² + 2x = -3 3(x² + (2x/3)) = -3 x² + (2x/3) = -1 x² + (2x/3) + (1/9) = -1 + (1/9) = -8/9 (x + (1/3))² = -8/9 x + 1/3= ±√(-8/9) = ±(2i/3)√2 x = (-1/3) ± (2i/3)√2 = (-1 ± 2i√2) / 3
👍 106 | 👎 -16 Kelleigh
The roots of the equation are the values where x = 0 3x^2 + 2x +3 = 0 Using Quadratic Formula to solve: a = 3, b = 2, c =3 x (roots) = [-2 ± √(2^2 - 4*3*3)] / (2 *3) x(roots) = [-2 ±√(4 -36)] / 6 x(roots) = [-2 ±√-32] / 6 x(roots) = [-2 ± √16 * √-2 ] / 6 x(roots) = [-2 ± 4* 2i] /6 x(roots) = -1/3 ± 8i/6
👍 99 | 👎 -24 Irmalinda
use the quadratic equation: http://www.daleheffernan.com/QuadraticFormula.jpg where y = ax^2 + bx + c so basically just move the equation around so that it equals 0. so 3x^2 +2x +3 = O then we have a= 3, b = 2, c = 3, and you sub those values in the quadratic equation giving you... No real roots because you can't square root a negative number. Are you sure your equation was right? If you're talking complex polynomials, the roots would be: x= 0.333 + 0.9428i or x= -0.333 -0.9428i
👍 92 | 👎 -32 Elfleda
3x ² + 2x + 3 = 0 x = [ - b ± √ ( b ² - 4 a c ) ] / 2 a x = [ - 2 ± √ ( 4 - 36 ) ] / 6 x = [ - 2 ± √ ( - 32 ) ] / 6 x = [ - 2 ± i √ ( 32 ) ] / 6 x = [ - 2 ± i 4√2 ] / 6 x = [ - 1 ± i 2√2 ] / 3
👍 85 | 👎 -40 Originally Answered: Square root equation.?