Solve x^2+2x-8=0

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Sujoy Das

JU-IITKGP & 10+ years experienced for boards, JEE,NEET,OLYMPIAD & 5-12

x=2 or -4
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BE(Electronics)

x^2+2x-8=0 then factorization splitting middle term we get x^2+4x-2x-8=0 Taking common we get x(x+4)-2(x+4)=0 after that we get (x-2) (x+4)=0 Hence we get x=2 or x= -4
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Founder, Accelerated Learning | 15+ Teaching Years

This is an equation of form ax^2 + bx + c Use the formula for calculating the roots as- x = (-b + (b - 4ac)^1/2 ) / 2a & x = (-b - (b - 4ac)^1/2 ) / 2a Substituting values into the variables you will get- x = (-2 + (2 - 4 x (1) x (-8)) ^1/2 ) / (2 x (1)) & x = (-2 - (2 - 4 x (1) x (-8)) ^1/2...
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This is an equation of form ax^2 + bx + c Use the formula for calculating the roots as- x = (-b + (b - 4ac)^1/2 ) / 2a & x = (-b - (b - 4ac)^1/2 ) / 2a Substituting values into the variables you will get- x = (-2 + (2 - 4 x (1) x (-8)) ^1/2 ) / (2 x (1)) & x = (-2 - (2 - 4 x (1) x (-8)) ^1/2 ) / (2 x (1)) On solving the above you will get the roots of the equation as- x = 2 & x = -4 This is the solution for the given quadratic equation. Hope you find this useful. read less
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Home Tutor

2 because only positive values will be selected
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Professional Tutor with 15 years of experience.

x(x+4)-2(x+4)=0 (x-2)(x+4)=0 x=2 or -4
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Deep Classes Nerul West Branch

x = -4 and x=2
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SME in Mathematics & Statistics

x^2 + 2x - 8=0 x^2 + 4x - 2x - 8=0 x(x + 4) - 2(x + 4) = 0 (x + 4)(x - 2) = 0 x = - 4 or 2
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Ms Zeba : The factors given are perfectly alright but the solution is incomplete. The factors (x+4)(x-2)=0,means x=-4 & x=2 give complete solution of given equation.
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BE, R V College Of Engineering, Bangalore.

x^2+4x-2x-8=0 =>x(x+4)-2(x+4)=0; =>(x+4)(x-2)=0 x=-4,x=2
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Teacher by Choice, not by profession

x^2+4x-2x-8=0 => x(x+4)-2(x+4)=0 => (x-2)(x+4)=0 => x= 2, -4
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