How do you solve using the quadratic formula method #2x^2 + 3x – 1 = 0#?

Question

Eli Assouline · Accepted Answer

#x=(-3+sqrt17)/4# or #x=(-3-sqrt17)/4# Quadratic formula which gives solution of #ax^2+bx+c=0# is #x=(-b+-sqrt(b^2-4ac))/(2a)#. In the equation #2x^2+3x–1=0#, #a=2#, #b=3# and #c=-1# and hence solution is #x=(-3+-sqrt(3^2-4xx2xx(-1)))/(2xx2)# or #x=(-3+-sqrt(9+8))/(2xx2)=-(3+-sqrt17)/4# i.e. #x=(-3+sqrt17)/4# or #x=(-3-sqrt17)/4#

Julia Adams · Answer

undefined To solve the quadratic equation $2x^2 + 3x - 1 = 0$ using the quadratic formula, where the equation is in the form $ax^2 + bx + c = 0$, follow these steps:

1. Identify $a$, $b$, and $c$ from the given equation:
   $a = 2$, $b = 3$, and $c = -1$.

2. Substitute the values of $a$, $b$, and $c$ into the quadratic formula:
   $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$.

3. Plug in the values:
   $x = \frac{{-3 \pm \sqrt{{3^2 - 4 \cdot 2 \cdot (-1)}}}}{{2 \cdot 2}}$.

4. Simplify under the square root:
   $x = \frac{{-3 \pm \sqrt{{9 + 8}}}}{{4}}$,
   $x = \frac{{-3 \pm \sqrt{{17}}}}{{4}}$.

5. The solutions are obtained by evaluating both the positive and negative square roots:
   $x_1 = \frac{{-3 + \sqrt{{17}}}}{{4}}$,
   $x_2 = \frac{{-3 - \sqrt{{17}}}}{{4}}$.