How to solve (4x)^(2)-5x+3=0

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (4x)^(2)-5x+3=0 | Equatix

Question

$${ \left(4x \right) }^{ 2 } -5x+3=0$$

Solve for x (complex solution)

$x=\frac{5+\sqrt{167}i}{32}\approx 0.15625+0.403838999i$
$x=\frac{-\sqrt{167}i+5}{32}\approx 0.15625-0.403838999i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Expand $\left(4x\right)^{2}$.

$$4^{2}x^{2}-5x+3=0$$

Calculate $4$ to the power of $2$ and get $16$.

$$16x^{2}-5x+3=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $16$ for $a$, $-5$ for $b$, and $3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 16\times 3}}{2\times 16}$$

Square $-5$.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 16\times 3}}{2\times 16}$$

Multiply $-4$ times $16$.

$$x=\frac{-\left(-5\right)±\sqrt{25-64\times 3}}{2\times 16}$$

Multiply $-64$ times $3$.

$$x=\frac{-\left(-5\right)±\sqrt{25-192}}{2\times 16}$$

Add $25$ to $-192$.

$$x=\frac{-\left(-5\right)±\sqrt{-167}}{2\times 16}$$

Take the square root of $-167$.

$$x=\frac{-\left(-5\right)±\sqrt{167}i}{2\times 16}$$

The opposite of $-5$ is $5$.

$$x=\frac{5±\sqrt{167}i}{2\times 16}$$

Multiply $2$ times $16$.

$$x=\frac{5±\sqrt{167}i}{32}$$

Now solve the equation $x=\frac{5±\sqrt{167}i}{32}$ when $±$ is plus. Add $5$ to $i\sqrt{167}$.

$$x=\frac{5+\sqrt{167}i}{32}$$

Now solve the equation $x=\frac{5±\sqrt{167}i}{32}$ when $±$ is minus. Subtract $i\sqrt{167}$ from $5$.

$$x=\frac{-\sqrt{167}i+5}{32}$$

The equation is now solved.

$$x=\frac{5+\sqrt{167}i}{32}$$ $$x=\frac{-\sqrt{167}i+5}{32}$$

Steps for Completing the Square

Expand $\left(4x\right)^{2}$.

$$4^{2}x^{2}-5x+3=0$$

Calculate $4$ to the power of $2$ and get $16$.

$$16x^{2}-5x+3=0$$

Subtract $3$ from both sides. Anything subtracted from zero gives its negation.

$$16x^{2}-5x=-3$$

Divide both sides by $16$.

$$\frac{16x^{2}-5x}{16}=-\frac{3}{16}$$

Dividing by $16$ undoes the multiplication by $16$.

$$x^{2}-\frac{5}{16}x=-\frac{3}{16}$$

Divide $-\frac{5}{16}$, the coefficient of the $x$ term, by $2$ to get $-\frac{5}{32}$. Then add the square of $-\frac{5}{32}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{16}x+\left(-\frac{5}{32}\right)^{2}=-\frac{3}{16}+\left(-\frac{5}{32}\right)^{2}$$

Square $-\frac{5}{32}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{16}x+\frac{25}{1024}=-\frac{3}{16}+\frac{25}{1024}$$

Add $-\frac{3}{16}$ to $\frac{25}{1024}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{16}x+\frac{25}{1024}=-\frac{167}{1024}$$

Factor $x^{2}-\frac{5}{16}x+\frac{25}{1024}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{5}{32}\right)^{2}=-\frac{167}{1024}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{32}\right)^{2}}=\sqrt{-\frac{167}{1024}}$$

Simplify.

$$x-\frac{5}{32}=\frac{\sqrt{167}i}{32}$$ $$x-\frac{5}{32}=-\frac{\sqrt{167}i}{32}$$

Add $\frac{5}{32}$ to both sides of the equation.

$$x=\frac{5+\sqrt{167}i}{32}$$ $$x=\frac{-\sqrt{167}i+5}{32}$$

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Example

Question

Solve for x (complex solution)

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square