How do you solve #8x+1=4x^2# by quadratic formula?

Question

Genesis Allen · Accepted Answer

#8x+1=4x^2#

#rarr8x=4x^2-1#

#rarr4x^2-1-8x=0#

In standard form:

#rarr4x^2-8x-1=0#

Now this is a Quadratic equation (in form #ax^2+bx^2+c=0#)

Use Quadratic formula:

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

In this case #a=4,b=-8,c=-1#

Substitute the values into the formula:

#rarrx=(-(-8)+-sqrt((-8)^2-4(4)(-1)))/(2(4))#

#rarrx=(8+-sqrt(64-(-16)))/8#

#rarrx=(8+-sqrt(64+16))/8#

#rarrx=(8+-sqrt(80))/8#

#rarrx=(8+-sqrt(16*5))/8#

#rarrx=(8+-4sqrt5)/8#

#rarrx=(2+-sqrt5)/2#

Nathan Axel · Answer

#(2+-sqrt(5))/(2)#

quadratic fomula

#ax^2+bx+c=0#

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

so #8x+1=4x^2#

#rarr4x^2-8x-1=0#

#rarr(-(-8)+-sqrt((-8)^2 -4(4)(-1)))/(2(4))#

#rarr(8+-sqrt(64 +16))/(8)#

#rarr(8+-sqrt(80))/(8)#

#rarr(8+-4sqrt(5))/(8)#

#rarr(2+-sqrt(5))/(2)#

Savannah Anderson · Answer

undefined To solve the quadratic equation 8x + 1 = 4x^2 using the quadratic formula:

1. First, rewrite the equation in standard form, which is ax^2 + bx + c = 0. So, move all terms to one side:
   4x^2 - 8x - 1 = 0

2. Identify the values of coefficients a, b, and c:
   a = 4, b = -8, c = -1

3. Plug these values into the quadratic formula:
   x = (-b ± √(b^2 - 4ac)) / (2a)

4. Substitute the values of a, b, and c into the quadratic formula:
   x = (-(-8) ± √((-8)^2 - 4 * 4 * (-1))) / (2 * 4)

5. Simplify inside the square root:
   x = (8 ± √(64 + 16)) / 8
   x = (8 ± √80) / 8

6. Further simplify:
   x = (8 ± √(16 * 5)) / 8
   x = (8 ± 4√5) / 8

7. Divide both the numerator and denominator by 4:
   x = (2 ± √5) / 2

So, the solutions to the equation 8x + 1 = 4x^2 by quadratic formula are:
   x = (2 + √5) / 2   and   x = (2 - √5) / 2