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GRE Algebra | Solving Quadratic Equations

  • Last Updated : 25 Apr, 2019

In algebra, a quadratic equation can be written in the form:

ax2 + bx + c = 0 

where x is the variable and a, b, c are the real numbers and a≠0. If a=0 then it will be a linear equation not quadratic because no second order term.

If quadratic equation has solution then it can be found by using the quadratic formula.

x = \frac{-b\pm \sqrt{b^2-4ac}}{2a} 
  • Example-1: Solve the quadratic equation for x,
    x2 + 10x -24 = 0 

    Solution: In the quadratic equation, we have,

    a=1, b=10 and c=-24 

    Therefore the quadratic formula yields



    x = \frac{-10\pm \sqrt{10^2-4(1)(-24)}}{2(1)} 
    
    x = \frac{-10\pm \sqrt{100+96}}{2} 
    
    x = \frac{-10\pm 14}{2}  

    Hence, two solutions for the above equations are:

    x = 4/2 = 2,
    And
    x = -24/2 = -12 
  • Example-2: Solve the quadratic equation using factorization,
    x2 + 2x - 15 = 0

    Solution: Given equation,

    x2 + 2x - 15 = 0 

    It can be factorize as,

    x2 -3x + 5x - 15 = 0
    x(x - 3) + 5(x - 3) = 0
    (x - 3)(x + 5) = 0 

    Hence, two solutions for x are:

    x = 3 and x = -5 
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