How do you solve #4+4x=2x^2# using the quadratic formula?

Question

Isaiah Anthony · Accepted Answer

#x=1+-sqrt(3)" "#Exact value

#x~~-0.732050....-> -0.73# to 2 decimal places

#x~~color(white)(-)2.732050...... ->+2.73 # to 2 decimal places

Subtract 4 and #4x# from both sides giving: #y=0=2x^2-4x-4# Consider the standard form of #y=ax^2+bx+c# That at #y=0# we have #x=(-b+-sqrt(b^2-4ac))/(2a)# In this case: #a=2"; "b=-4"; "c=-4# so by substitution we have: #x=(+4+-sqrt((-4)^2-4(2)(-4)))/(2(2))# #x=1+-sqrt(48)/4# ............................................................... Note that #sqrt(48)/4# is the same as #sqrt(48/4^2)=sqrt(3)# ..................................................................... #x=1+-sqrt(3)" "#Exact value #x~~-0.732050....-> -0.73# to 2 decimal places #x~~color(white)(-)2.732050...... ->+2.73 # to 2 decimal places

Eli Assouline · Answer

undefined To solve the equation $4 + 4x = 2x^2$ using the quadratic formula, first rearrange the equation into the standard form $ax^2 + bx + c = 0$. Then, identify the coefficients $a$, $b$, and $c$. After that, substitute these values into the quadratic formula $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, and solve for $x$.

For the equation $4 + 4x = 2x^2$, we have $a = 2$, $b = -4$, and $c = -4$.

Substituting these values into the quadratic formula, we get:

$$x = \frac{{-(-4) \pm \sqrt{{(-4)^2 - 4(2)(-4)}}}}{{2(2)}}$$

Simplify:

$$x = \frac{{4 \pm \sqrt{{16 + 32}}}}{{4}}$$

$$x = \frac{{4 \pm \sqrt{{48}}}}{{4}}$$

$$x = \frac{{4 \pm 4\sqrt{{3}}}}{{4}}$$

$$x = \frac{{1 \pm \sqrt{{3}}}}{1}$$

Therefore, the solutions are $x = \frac{{1 + \sqrt{{3}}}}{2}$ and $x = \frac{{1 - \sqrt{{3}}}}{2}$.