Solve the quadratic equation 5×2 + x + 1 = 0.
The correct answer and explanation is:
Let’s solve the quadratic equation:
5×2+x+1=05x^2 + x + 1 = 0
Step 1: Identify coefficients
The general quadratic equation is:
ax2+bx+c=0ax^2 + bx + c = 0
Here,
a=5a = 5,
b=1b = 1,
c=1c = 1.
Step 2: Use the quadratic formula
The quadratic formula to solve for xx is: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
Step 3: Calculate the discriminant
Calculate the discriminant Δ\Delta: Δ=b2−4ac=(1)2−4(5)(1)=1−20=−19\Delta = b^2 – 4ac = (1)^2 – 4(5)(1) = 1 – 20 = -19
Since the discriminant is negative (−19-19), this equation has no real roots, but two complex (imaginary) roots.
Step 4: Find the roots
Substitute values into the quadratic formula: x=−1±−192×5=−1±−1910x = \frac{-1 \pm \sqrt{-19}}{2 \times 5} = \frac{-1 \pm \sqrt{-19}}{10}
Recall −19=19i\sqrt{-19} = \sqrt{19}i, where ii is the imaginary unit (i2=−1i^2 = -1).
So, x=−1±i1910x = \frac{-1 \pm i\sqrt{19}}{10}
Final answer:
x=−110+1910iorx=−110−1910ix = \frac{-1}{10} + \frac{\sqrt{19}}{10}i \quad \text{or} \quad x = \frac{-1}{10} – \frac{\sqrt{19}}{10}i
Explanation (300 words):
The quadratic equation 5×2+x+1=05x^2 + x + 1 = 0 is a polynomial equation of degree 2, where xx is the variable we want to solve for. Quadratic equations can have two real roots, one real root (when the roots are equal), or two complex roots, depending on the discriminant b2−4acb^2 – 4ac.
The discriminant is key to determining the nature of the roots:
- If it is positive, the equation has two distinct real roots.
- If it is zero, there is exactly one real root (a repeated root).
- If it is negative, the roots are complex conjugates, involving the imaginary unit ii.
In this case, the discriminant was −19-19, a negative number, which means the parabola described by the quadratic function does not cross the x-axis. Instead, the solutions lie in the complex plane.
Using the quadratic formula, which is a universal method to solve any quadratic equation, we substituted the coefficients and found the roots as complex numbers: x=−110±1910ix = \frac{-1}{10} \pm \frac{\sqrt{19}}{10}i
These roots are conjugates of each other, which is common for polynomials with real coefficients when the discriminant is negative.
The real part −110-\frac{1}{10} shifts the solutions along the real axis, while the imaginary parts ±1910i\pm \frac{\sqrt{19}}{10}i represent the vertical distance from the real axis in the complex plane.
In summary, the quadratic equation has no real solutions but two complex roots that can be expressed in standard form as shown above. This understanding is fundamental in algebra and complex number theory, ensuring every quadratic equation has two solutions in the complex number system.