Find the zeros of m(x)=-5x^(2)+50x-135 using quadratic formula with complex numbers

Question

asked Jan 16, 2024 23.7k views

1 Answer

Peter Turner · Answer 1 · 2024-01-20T14:02:58+0000

Answer:

$x=5\pm √(2)\:i$

Explanation:

The quadratic formula is a mathematical expression used to find the solutions of a quadratic equation of the form ax² + bx + c = 0:

$\boxed{\begin{array}{l}\underline{\sf Quadratic\;Formula}\\\\x=(-b \pm √(b^2-4ac))/(2a)\\\\\textsf{when} \;ax^2+bx+c=0 \\\end{array}}$

In the case of m(x) = -5x² + 50x - 135:

Substitute the values of a, b and c into the quadratic formula:

$x=(-50\pm√(50^2-4(-5)(-135)))/(2(-5))$

Simplify:

$x=(-50\pm√(2500-2700))/(-10)$

$x=(-50\pm√(-200))/(-10)$

Rewrite 200 as the product of 10², 2 and -1:

$x=(-50\pm√(10^2\cdot2\cdot(-1)))/(-10)$

$\textsf{Apply the radical rule:} \quad √(ab)=\sqrt{\vphantom{b}a}√(b)$

$x=(-50\pm√(10^2)√(2)√(-1))/(-10)$

$\textsf{Apply the radical rule:} \quad √(a^2)=a, \quad a \geq 0$

$x=(-50\pm10√(2)√(-1))/(-10)$

$\textsf{Apply the imaginary number rule:} \quad √(-1)=i$

$x=(-50\pm10√(2)\:i)/(-10)$

Simplify by dividing the numerator and denominator by the common factor of -10:

$x=5\pm √(2)\:i$

Therefore, the two solutions of the quadratic equation are: