Algebra · real student question

Solve the inequality x2 + x - 1 is greater than or equal to 0.

Question

Solve for xx:

x2+x10x^2+x-1\ge 0

Step-by-step solution

  1. Locate the roots with the quadratic formula. With a=1a=1, b=1b=1, c=1c=-1,

    Δ=124(1)(1)=1+4=5x=1±52\Delta=1^2-4(1)(-1)=1+4=5\quad\Rightarrow\quad x=\frac{-1\pm\sqrt5}{2}

    No integer pair multiplies to 1-1 and adds to 11, so factoring was never going to work.

  2. Recognise the numbers. 1+52=0.6180340\frac{-1+\sqrt5}{2}=0.6180340 and 152=1.6180340\frac{-1-\sqrt5}{2}=-1.6180340 — these are the golden-ratio conjugates, satisfying φ2=φ+1\varphi^2=\varphi+1 which is precisely x2+x1=0x^2+x-1=0 rearranged. Their product is 1-1 and their sum is 1-1, matching c/ac/a and b/a-b/a.

  3. Apply the opening direction. The leading coefficient is positive, so the parabola opens upward and the expression is 0\ge 0 outside the interval between the roots:

    x152orx1+52x\le\frac{-1-\sqrt5}{2}\qquad\text{or}\qquad x\ge\frac{-1+\sqrt5}{2}

  4. Include the endpoints. The inequality is non-strict (\ge), and at each root the expression equals exactly 00, so both boundary values belong to the solution set — closed brackets, not open ones.

  5. Verify with test points. At x=2x=-2: 421=104-2-1=1\ge 0 ✓. At x=0x=0: 1-1, which fails, correctly excluding the middle ✓. At x=1x=1: 1+11=101+1-1=1\ge 0 ✓. So the solution is

    (,152][1+52,)\left(-\infty,\tfrac{-1-\sqrt5}{2}\right]\cup\left[\tfrac{-1+\sqrt5}{2},\infty\right)

Answer

x1521.6180orx1+520.6180x\le\frac{-1-\sqrt5}{2}\approx -1.6180\quad\text{or}\quad x\ge\frac{-1+\sqrt5}{2}\approx 0.6180

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