Quadratic formula - need help!?
I'm supposed to apply the quadratic formula on this: 4w^2 + 50w - 100 = 0 But I'm stuck at this point: w^2 + 12.5 - 25 = 0 -12.5 ± √12.5^2 - 4 (-25)/2 -12.5 ± √12.5^2 + 100/2 -12.5 ± √256.25/2 I don't know how to get rid of the radical :/ I'm sure I did something wrong and was supposed to get another number but I don't see where the problem is. Also, could you explain in more detail instead of just writing the answer please? It would help me a lot.
I need help with quadratic formula!!?
2 = -10/x - 5/x^2 x^2(2) = x^2(-10/x - 5/x^2) 2x^2 = -10x - 5 2x^2 + 10x + 5 = 0 x = [-b ±√(b^2 - 4ac)]/2a a = 2 b = 10 c = 5 x = [-10 ±√(100 - 40)]/4 x = [-10 ±√60]/4 x = [-10 ±√(2^2 * 15)]/4 x = [-10 ±2√15]/4 x = [-5 ±√15]/2 ∴ x = [-5 ±√15]/2
Need help quadratic formula and some equation?
Problem 1: x ^2 - 2x + 5 = 0 The quadratic formula is, x = [ - b + or - sqrt ( b ^ 2 - 4 a c ) ] / 2 a Here, we can take disc = b ^2 - 4 a c If disc = 0 then throots are same. If disc > 0 then the given equation has real roots. If disc < 0 then the given equation has no real roots. so, we can say that it has imaginary roots. In our equation a = 1 , b = -2 and c = 5 Plug the values in the formula we get, x = [ - ( - 2 ) + or - sqrt ( (-2 ) ^2 - 4 * 1 * 5 ) ] / 2(1) = [ 2 + or - sqrt (4 - 20 ) ] / 2 = [ 2 + or - sqrt ( -16 ) ] / 2 Here, the disc = -16 < 0. So, the given equation has no real roots. So, the imaginary roots are, x = (2 + or - 4i) / 2 , i - imaginary. = 2 ( 1 + or - 2i ) / 2 cancel out 2. So, we have x = 1 + or - 2i Therefore, the roots are, x = 1 + 2i and x = 1 - 2i Problem 2 : The given equation is 5 m^2 + 3m = 2 we can write this as 5m ^ 2 + 3m - 2 = 0 We can solve this equation by factoring, Here, 5 * - 2 = -10 So, we have to find two numbers whose product is equal to -10 and their sum is equal to 3. Such numbers are 5 * - 2 = -10 and 5 + (-2) = 3 Therefore, The equation becomes, 5 m ^2 + 3m - 2 = 0 implies that . 5 m ^2 + 5m - 2m - 2 = 0 5 m ( m + 1 ) - 2 ( m + 1 ) = 0 ( m + 1 ) ( 5 m - 2 ) = 0 (m+1) = 0 or ( 5m - 2 ) = 0 This implies that , m = -1 and 5 m = 2 m = 2 / 5 Hence, the roots are, m = -1 and m = 2 / 5. ======================================... For ore assistance please visit www.classof1.com. Classof1 - leading web tutoring and homework help. Toll free - 1877-252-7763
Need help with the quadratic formula!?
Need help with this quadratic formula problem. I think I did it right but I just want to make sure. This is for my college math placement test so just need to make sure I understand everything so that I can answer multiple problems of this style. This is the question and my answer please tell me if I am correct and if I am wrong please write an explanation step by step. Solve 3x^2-2x+27=0 using the quadratic formula x=-b±√b^2-4ac/2a express the answer in simplest radical the answer I have is x=2±3√5/6 is that correct thanks for your help
I need help with solving using the quadratic formula?
question type a million : For this equation x^2 - 7*x - a million = - 7 , answer right here questions : A. locate the roots using Quadratic formulation ! B. Use factorization to locate the basis of the equation ! C. Use polishing off the sq. to locate the basis of the equation ! answer type a million : First, we could desire to consistently show equation : x^2 - 7*x - a million = - 7 , right into a*x^2+b*x+c=0 sort. x^2 - 7*x - a million = - 7 , flow each and every thing interior the terrific hand area, to the left hand area of the equation <=> x^2 - 7*x - a million - ( - 7 ) = 0 , it incredibly is the comparable with <=> x^2 - 7*x - a million + ( 7 ) =0 , now open the bracket and we get <=> x^2 - 7*x + 6 = 0 The equation x^2 - 7*x + 6 = 0 is already in a*x^2+b*x+c=0 sort. In that kind, we can actual derive that the fee of a = a million, b = -7, c = 6. 1A. locate the roots using Quadratic formulation ! Use the formulation, x1 = (-b+sqrt(b^2-4*a*c))/(2*a) and x2 = (-b-sqrt(b^2-4*a*c))/(2*a) We had understand that a = a million, b = -7 and c = 6, we could desire to subtitute a,b,c interior the abc formulation, with thos values. Which produce x1 = (-(-7) + sqrt( (-7)^2 - 4 * (a million)*(6)))/(2*a million) and x2 = (-(-7) - sqrt( (-7)^2 - 4 * (a million)*(6)))/(2*a million) it incredibly is the comparable with x1 = ( 7 + sqrt( 40 9-24))/(2) and x2 = ( 7 - sqrt( 40 9-24))/(2) Which make x1 = ( 7 + sqrt( 25))/(2) and x2 = ( 7 - sqrt( 25))/(2) So we get x1 = ( 7 + 5 )/(2) and x2 = ( 7 - 5 )/(2) So we've the solutions x1 = 6 and x2 = a million 1B. Use factorization to locate the basis of the equation ! x^2 - 7*x + 6 = 0 ( x - 6 ) * ( x - a million ) = 0 The solutions are x1 = 6 and x2 = a million 1C. Use polishing off the sq. to locate the basis of the equation ! x^2 - 7*x + 6 = 0 ,divide the two area with a million Then we get x^2 - 7*x + 6 = 0 , all of us understand that the coefficient of x is -7 we could desire to consistently use the very undeniable fact that ( x + q )^2 = x^2 + 2*q*x + q^2 , and anticipate that q = -7/2 = -3.5 So we've make the equation into x^2 - 7*x + 12.25 - 6.25 = 0 that is grew to become into ( x - 3.5 )^2 - 6.25 = 0 So we can get (( x - 3.5 ) - 2.5 ) * (( x - 3.5 ) + 2.5 ) = 0 through using the associative regulation we get ( x - 3.5 - 2.5 ) * ( x - 3.5 + 2.5 ) = 0 And it incredibly is the comparable with ( x - 6 ) * ( x - a million ) = 0 So we've been given the solutions as x1 = 6 and x2 = a million
I need help with some quadratic equations?
One common time era that seems in lots of cases is "equations." for many, this may be a buzz be conscious that's recognizable as a math time era yet fairly information the time era could sometime get away human beings. the two maximum common equations are linear equations and quadratic equations yet, of direction, in straight forward terms giving an equation a popularity would not inevitably clarify what that's. So, a clearer definition of what precisely the two maximum common equations - linear and quadratic equations - is needed. A linear equation refers to a particular equation that's graphed on a promptly line. additionally, a linear equation possesses on the line one variable that's frequently stated as "X" and "X" will constantly be of a level that's a million at maximum. (that's, there are no exponents; yet once you're searching for for exponents then wait and notice because of the fact we can get to them presently!) a typical occasion of a linear equation may be 1x + 2 = 3. of course, x might equivalent a million in this actual occasion and this is discovered by using in straight forward terms employing slightly algebra on the equation to ensure X. 3 minus 2 equals a million. as a result, X could equivalent one as a million x a million equals one. And, nope, no longer all linear equations are that ordinary as they arrive as complicated as 6(x + 3) = 24 (x +0), however the common ingredient of putting apart x to discover the respond would not substitute. A quadratic equation is in basic terms diverse from a linear equation in one understand: a number of of of the figures is squared. (The be conscious quadratic derives from the Latin be conscious for squared) the common style of a quadratic equation is ax2 + bx + c = eleven. In considered one of those equation, if a = a million, b = 2 and c = 3 then X could equivalent 2. all of us comprehend this because of the fact 2 squared is 4 and four x a million = 4. 2 x 2 = 4. As with a linear equation, there may be greater complicated variations of a quadratic equation yet in basic terms with the ordinary and sophisticated linear equations common algebraic operations can yield the main recommendations-blowing answer.
Need Help With Maths! Quadratic Equations!?
x2-9x+14: simplify the equation: (x-7)(x-2)...factors of +14 that can add up to -9 then transpose: x=7, x=2 this is an easy way of solving quadratic equations that start with x2.(x squared) it's much easier to understand it if you ask a classmate or friend of yours(preferably the smart ones or someone who does understand) to help you.out. Hope this helped you! :D