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Topic: **Problem solving by quadratic equations****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

July 16, 2019 / By Arran

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.

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There are a wide variety of forms for the equation of a line. Because you are using m, the letter representing gradient, and b, the letter often used for the intercept, you probably want gradient-intercept form. y = mx + b where m is the gradient and b is the y-intercept. 1. m = -2/5, b = 3 y = (-2/5)x + 3 2. m = 0, b = 4 y = (0)x + 4 = y = 4. For other ways of writing it, see the wikipedia source referenced below from 1.1 - 'Linear equations in two variables' - to 1.1.9 - 'Normal form'. It starts by introducing the gradient-intercept form, although it doesn't name it. Note that it is a variant of the 'Point-slope' form; the point, however, is necessarily the y-intercept in the gradient-intercept form. The form is called 'gradient-intercept', because you have the line described as a function of y in two steps: 'y = ...' [The line represents a function of y where ...] 'mx ...' [The slope, or gradient, of the line is proportional in a ratio defined by m] '+ b' [And the y-intercept of the line is (0, b)] So the information, in order, is gradient, then intercept... gradient-intercept. We could equally write a gradient-intercept form as a function of x, x = my + b, but functions are almost always written in terms of y, and this is assumed when referring to a gradient-intercept form.

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.

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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

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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

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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

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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

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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

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Hi, To solve equations with radicals: 1) Get a term with a radical alone on one side of the equation. If there is a number outside multiplied times the radical, divide by that number if it goes evenly into the other terms. 2) With the radical alone, raise both SIDES to a power equal to the radical's index number. For square roots, square both sides, For cube roots, cube both sides, etc. Notice you square or cube the entire side. You do NOT square or cube each term separately, but instead you square or cube the entire SIDE. 3) Combine like terms to simplify the new equation. If there is still a radical, repeat steps 1 and 2. If not solve the equation. 4) When you get 1 or more answers, they may include extraneous answers that won't check in the original problem. So, you MUST plug each answer back into the original equation to see if they are really true, good answers. If they don't check, discard them. If there are 2 answers both could work, only 1 could work or neither could work. So ALWAYS check answers. Here's your problem: ..____.....___ √2x+2.-.√x+2.=.1 Get a radical alone. ..____...........___ √2x+2.=.1.-.√x+2 Square both SIDES, not term by term. ...____..............___ (√2x+2)².=.(1.-.√x+2)² On the left side, the square and square root cancel each other out. ...................___ 2x+2.=.(1.-.√x+2)² ........___........___ (1.-.√x+2)(1.-.√x+2) = 2x + 2 ..........___ 1 - 2√x+2 + x + 2 = 2x + 2 Simplify by combining like terms. Get the radical alone. ......___ -2√x+2 = x - 1 Now square both sides and solve. .......___ (-2√x+2)² = (x - 1)² 4(x+2) = x² - 2x + 1 4x + 8 = x² - 2x + 1 x² - 6x - 7 = 0 (x - 7)(x + 1) = 0 x - 7 = 0 x = 7 x + 1 = 0 x = -1 Check these answers in the original equation to see if they check. Let x = 7 ..____.....___ √2x+2.-.√x+2.=.1 ..______......___ √2(7)+2...-.√7+2.=.1 4 - 3 = 1 so x = 7 checks. Let x = -1 ..____.....___ √2x+2.-.√x+2.=.1 ..______......____ √2(-1)+2...-.√-1+2.=.1 0 - 1 ≠ 1 so x = -1 does NOT check and is discarded. x = 7 is the only solution. I hope that helps!! :-)

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