Algebra · real student question

Solve the inequality 2x2 + x - 1 > 0.

Question

Solve for xx:

2x2+x1>02x^2+x-1>0

Step-by-step solution

  1. Factor the quadratic. Split the middle term using two numbers with product 2(1)=22\cdot(-1)=-2 and sum +1+1, namely +2+2 and 1-1:

    2x2+2xx1=2x(x+1)1(x+1)=(2x1)(x+1)2x^2+2x-x-1=2x(x+1)-1(x+1)=(2x-1)(x+1)

    Factoring first is worth the effort: the roots then appear without any square roots.

  2. Read off the roots. (2x1)(x+1)=0(2x-1)(x+1)=0 gives x=12x=\frac12 and x=1x=-1. These split the line into (,1)(-\infty,-1), (1,12)\left(-1,\frac12\right) and (12,)\left(\frac12,\infty\right).

  3. Predict the answer from the shape. The leading coefficient 2>02>0, so the parabola opens upward: it is positive outside its two roots and negative strictly between them. That already gives the shape of the answer.

  4. Confirm with one test point per interval.

    x=2: (5)(1)=5>0x=0: (1)(1)=1<0x=1: (1)(2)=2>0x=-2:\ (-5)(-1)=5>0\qquad x=0:\ (-1)(1)=-1<0\qquad x=1:\ (1)(2)=2>0

    Matching the prediction: the outer intervals work, the middle one does not. Structurally, the product is positive when both factors share a sign.

  5. Handle the endpoints. At x=1x=-1 and x=12x=\frac12 the expression is exactly 00, and 0>00>0 is false, so the inequality is strict at both ends:

    x(,1)(12,)x\in(-\infty,-1)\cup\left(\tfrac12,\infty\right)

Answer

x<1orx>12,x(,1)(12,)x<-1\quad\text{or}\quad x>\frac12,\qquad x\in(-\infty,-1)\cup\left(\tfrac12,\infty\right)

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