How do you solve #5x^2 - 2x +4 = 0# by quadratic formula?

Question

Nathan Axel · Accepted Answer

#x=(1+isqrt19)/5,##x=(1-isqrt19)/5#

Solve by quadratic formula:

#5x^2-2x+4=0# is a quadratic equation in standard form: #ax^2+bx+c#, where #a=5#, #b=-2#, and #c=4#.

quadratic formula:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Substitute the values for #a,b, and c# into the formula.

#x=(-(-2)+-sqrt((-2)^2-4*5*4))/(2*5)#

Simplify.

#x=(2+-sqrt(4-80))/10#

#x=(2+-sqrt(-76))/10#

Prime factorize the number of the square root.

#x=(2+-isqrt(2xx2xx19))/10#

Simplify.

#x=(2+-2isqrt19)/10#

Reduce.

#x=(1+-isqrt19)/5#

Solutions for #x#.

#x=(1+isqrt19)/5,##x=(1-isqrt19)/5#

Isaiah Anthony · Answer

undefined To solve the quadratic equation $5x^2 - 2x + 4 = 0$ using the quadratic formula, follow these steps:

1. Identify the coefficients $a$, $b$, and $c$ in the quadratic equation $ax^2 + bx + c = 0$.
   
2. Apply the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$.

3. Substitute the values of $a$, $b$, and $c$ into the quadratic formula.

4. Calculate the discriminant ($b^2 - 4ac$) to determine the nature of the roots:
   - If the discriminant is positive, there are two real roots.
   - If the discriminant is zero, there is one real root (a repeated root).
   - If the discriminant is negative, there are two complex roots.

5. Substitute the values of $a$, $b$, and $c$ into the quadratic formula and calculate the roots by performing the arithmetic operations.