4x²-3x+2=0

asked by guest
on Mar 17, 2025 at 2:39 am



You asked:

Solve the equation 4x23x+2=04 {x}^{2} - 3 x + 2 = 0 for the variable xx.

MathBot Answer:

x=38+i2380.375+0.59947894ix=38i2380.3750.59947894i\begin{aligned}x &= \frac{3}{8} + \frac{i \sqrt{23}}{8} \approx 0.375 + 0.59947894 i\\x &= \frac{3}{8} - \frac{i \sqrt{23}}{8} \approx 0.375 - 0.59947894 i\end{aligned}

Solve by quadratic formula:

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=4a=4, b=3b=-3, and c=2c=2.

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=(3)2162=23<0b^{2}-4ac = \left(-3\right)^{2} - 16 \cdot 2=-23 < 0 The discriminant is less than zero, so this quadratic equation has two solutions with non-zero imaginary part.

The two solutions are: x=(1)(3)+2324=38+23i80.375+0.59947894ix = \frac{\left(-1\right) \left(-3\right) + \sqrt{-23}}{2 \cdot 4} = \frac{3}{8} + \frac{\sqrt{23} i}{8} \approx 0.375 + 0.59947894 i x=(1)(3)2324=3823i80.3750.59947894ix = \frac{\left(-1\right) \left(-3\right) - \sqrt{-23}}{2 \cdot 4} = \frac{3}{8} - \frac{\sqrt{23} i}{8} \approx 0.375 -0.59947894 i


ii is the imaginary unit, defined as i2=1i^2 = -1.