# How do you solve  4x^2+3=8x?

Mar 22, 2018

The solutions to the equation are $x = \frac{1}{2}$ and $x = \frac{3}{2}$.

#### Explanation:

We know that this is a quadratic since it has an ${x}^{2}$ term in it.

To solve quadratics, first set one side equal to $0$:

$4 {x}^{2} + 3 = 8 x$

$4 {x}^{2} + 3 - 8 x = 0$

$4 {x}^{2} - 8 x + 3 = 0$

Now, we have to find two numbers that multiply to $12$ (the $a$ term times the $c$ term) and add up to $- 8$ (the $b$ term).

After some experimentation, you will find that these two numbers are $- 6$ and $- 2$.

Now, split up the $x$ terms into these two numbers, then factor the first two terms and the last two terms separately, and lastly, combine them:

$4 {x}^{2} - 6 x - 2 x + 3 = 0$

$\textcolor{red}{2 x} \left(2 x - 3\right) - 2 x + 3 = 0$

$\textcolor{red}{2 x} \left(2 x - 3\right) - \textcolor{b l u e}{1} \left(2 x - 3\right) = 0$

$\left(\textcolor{red}{2 x} - \textcolor{b l u e}{1}\right) \left(2 x - 3\right) = 0$

Now, solve for when each factor equals $0$, and those will be the solutions:

color(white){color(black)( (2x-1=0,qquadqquad2x-3=0), (2x=1,qquadqquad2x=3), (x=1/2,qquadqquadx=3/2):}

Those are the answers. We can check out answers by graphing the quadratic $4 {x}^{2} - 8 x + 3$ on a calculator and seeing where it crosses the $x$-axis (it should cross when $x = \frac{1}{2}$ and also when $x = \frac{3}{2}$):

graph{4x^2-8x+3 [-0.5, 2.5, -1.29, 0.419]}

It does, so our answers are correct. Hope this helped!

Mar 22, 2018

$x = \frac{1}{2}$ and $x = \frac{3}{2}$

#### Explanation:

Solve the equation by the new Transforming Method (Socratic Search).

$y = 4 {x}^{2} - 8 x + 3 = 0$

Transformed equation:

$y ' = {x}^{2} - 8 x + 12$.

Proceeding. Find $2$ real roots of $y '$, then, divide them by $a = 4$.

Find $2$ real roots knowing the sum $\left(- b = 8\right)$ and the product
$\left(a c = 12\right)$. They are: $2$ and $6$.

The $2$ real roots of $y$ are:

${x}_{1} = \frac{2}{a} = \frac{2}{4} = \textcolor{b l u e}{\frac{1}{2}}$ and ${x}_{2} = \frac{6}{a} = \frac{6}{4} = \textcolor{b l u e}{\frac{3}{2}}$.