what are the solutions of x2-4x+5=0
The correct answer and explanation is:
Let’s solve the quadratic equation:
x2−4x+5=0x^2 – 4x + 5 = 0
Step 1: Identify coefficients
The quadratic equation is in standard form:
ax2+bx+c=0ax^2 + bx + c = 0
Here,
- a=1a = 1
- b=−4b = -4
- c=5c = 5
Step 2: Calculate the discriminant
The discriminant Δ\Delta is given by: Δ=b2−4ac\Delta = b^2 – 4ac
Substitute the values: Δ=(−4)2−4(1)(5)=16−20=−4\Delta = (-4)^2 – 4(1)(5) = 16 – 20 = -4
Step 3: Analyze the discriminant
- If Δ>0\Delta > 0, two real and distinct solutions.
- If Δ=0\Delta = 0, one real solution (a repeated root).
- If Δ<0\Delta < 0, two complex conjugate solutions.
Since Δ=−4\Delta = -4 (less than zero), the equation has two complex conjugate solutions.
Step 4: Use the quadratic formula
The quadratic formula is: x=−b±Δ2ax = \frac{-b \pm \sqrt{\Delta}}{2a}
Since Δ=−4\Delta = -4, we write: x=−(−4)±−42×1=4±−42x = \frac{-(-4) \pm \sqrt{-4}}{2 \times 1} = \frac{4 \pm \sqrt{-4}}{2}
Recall that −4=4×−1=2i\sqrt{-4} = \sqrt{4} \times \sqrt{-1} = 2i, where ii is the imaginary unit (i2=−1i^2 = -1).
So, x=4±2i2=2±ix = \frac{4 \pm 2i}{2} = 2 \pm i
Final solutions:
x=2+iandx=2−ix = 2 + i \quad \text{and} \quad x = 2 – i
Explanation:
The quadratic equation x2−4x+5=0x^2 – 4x + 5 = 0 has no real solutions because the discriminant is negative. This means the parabola y=x2−4x+5y = x^2 – 4x + 5 does not intersect the x-axis. Instead, its solutions are complex numbers involving the imaginary unit ii.
Complex solutions always come in conjugate pairs when the coefficients of the quadratic are real numbers. The solutions 2+i2 + i and 2−i2 – i reflect this conjugate pair nature.
The real part of both solutions is 2, indicating a “center” of the parabola along the x-axis, while the imaginary part ±i\pm i represents the distance from the real axis in the complex plane.
Quadratic equations like this often appear in algebra, engineering, and physics when modeling systems where the solutions represent oscillations or waves rather than straightforward real values.
If you graph this quadratic, the curve will open upwards with a vertex at x=2x = 2, y=1y = 1, and never touch the x-axis, which aligns with the fact that it has complex solutions.