Question

question
what are the roots of the equation (4x^2 + 23 = 19x) in simplest (a + bi) form?
answer
attempt 1 out of 2
+ additional solution - no solution
(x =) submit answer
(sqrt{quad}) (pm)

Explanation:

Step1: Rewrite the equation in standard form

First, we rewrite the quadratic equation \(4x^{2}+23 = 19x\) in the standard form \(ax^{2}+bx + c=0\). Subtract \(19x\) from both sides:
\(4x^{2}-19x + 23=0\)
Here, \(a = 4\), \(b=- 19\), and \(c = 23\).

Step2: Use the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). First, we calculate the discriminant \(\Delta=b^{2}-4ac\).
Substitute \(a = 4\), \(b=-19\), and \(c = 23\) into the discriminant formula:
\(\Delta=(-19)^{2}-4\times4\times23\)
\(=361-368\)
\(=- 7\)

Step3: Substitute into the quadratic formula

Now, substitute \(a = 4\), \(b=-19\), and \(\Delta=-7\) into the quadratic formula:
\(x=\frac{-(-19)\pm\sqrt{-7}}{2\times4}=\frac{19\pm i\sqrt{7}}{8}\) (since \(\sqrt{-7}=i\sqrt{7}\))

Answer:

The roots of the equation are \(x=\frac{19}{8}+\frac{\sqrt{7}}{8}i\) and \(x=\frac{19}{8}-\frac{\sqrt{7}}{8}i\)