n^2+n-4*10^18=0

asked by guest
on Apr 10, 2025 at 4:21 am



You asked:

Solve the equation n2+n41018=0{n}^{2} + n - 4 \cdot {10}^{18} = 0 for the variable nn.

MathBot Answer:

n=12+1600000000000000000122.0109n=121600000000000000000122.0109\begin{aligned}n &= - \frac{1}{2} + \frac{\sqrt{16000000000000000001}}{2} \approx 2.0 \cdot 10^{9}\\n &= - \frac{1}{2} - \frac{\sqrt{16000000000000000001}}{2} \approx -2.0 \cdot 10^{9}\end{aligned}

Solve by quadratic formula:

Simplify the left hand side of the equation into standard form: 41018+n2+n=0n2+n41000000000000000000=0n2+n4000000000000000000=0n2+n4000000000000000000=0\begin{aligned}- 4 \cdot 10^{18} + n^{2} + n &= 0\\n^{2} + n - 4 \cdot 1000000000000000000 &= 0\\n^{2} + n -4000000000000000000 &= 0\\n^{2} + n - 4000000000000000000 &= 0\end{aligned}

Given a quadratic equation ax2+bx+c=0a x^{2} + b x + c = 0, where aa, bb, cc are constants and a0a \ne 0, the solutions are given by the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} In this case a=1a=1, b=1b=1, and c=4000000000000000000c=-4000000000000000000.

The discriminant is the quantity under the square root sign in the quadratic formula, and its sign determines the number of solutions to the quadratic equation when the coefficients are real. The discriminant is:b24ac=124(4000000000000000000)=16000000000000000001>0b^{2}-4ac = 1^{2} - 4 \left(-4000000000000000000\right)=16000000000000000001 > 0 The discriminant is greater than zero, so this quadratic equation has two real solutions.

The two solutions are: n=(1)1+1600000000000000000121=12+1600000000000000000122.0109n = \frac{\left(-1\right) 1 + \sqrt{16000000000000000001}}{2 \cdot 1} = - \frac{1}{2} + \frac{\sqrt{16000000000000000001}}{2} \approx 2.0 \cdot 10^{9} n=(1)11600000000000000000121=121600000000000000000122.0109n = \frac{\left(-1\right) 1 - \sqrt{16000000000000000001}}{2 \cdot 1} = - \frac{1}{2} - \frac{\sqrt{16000000000000000001}}{2} \approx -2.0 \cdot 10^{9}