x - 3 = x^2 + 29を解く: 実数解なし

Problem

Solve $x - 3 = x^2 + 29$.

Step 1: Move the $x$ Term

We want all the $x$ pieces on the same side.

Subtract $x$ from both sides:

$x - 3 - x = x^2 + 29 - x$

This gives:

$-3 = x^2 - x + 29$

Step 2: Move the Number Term

Now we want one side to be $0$.

Add $3$ to both sides:

$-3 + 3 = x^2 - x + 29 + 3$

This gives:

$0 = x^2 - x + 32$

So the standard form is:

$x^2 - x + 32 = 0$

Step 3: Name the Three Coefficients

The standard form is:

$ax^2 + bx + c = 0$

For:

$x^2 - x + 32 = 0$

the three numbers are:

$a = 1$

$b = -1$

$c = 32$

Step 4: Build the Discriminant

The discriminant is the part under the square root in the quadratic formula:

$b^2 - 4ac$

Substitute $a = 1$, $b = -1$, and $c = 32$:

$(-1)^2 - 4(1)(32)$

Now simplify:

$1 - 128 = -127$

So the discriminant is:

$-127$

Step 5: Read the Answer Type

The discriminant is negative:

$-127 < 0$

That means the square root part is not a real number.

So this equation has no real-number solution.

Step 6: Write the Complex Solutions

Use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values:

$x = \frac{-(-1) \pm \sqrt{-127}}{2(1)}$

Simplify:

$x = \frac{1 \pm \sqrt{-127}}{2}$

Since $\sqrt{-127} = i\sqrt{127}$, the solutions are:

$x = \frac{1 \pm i\sqrt{127}}{2}$