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Algebra

Löse x - 3 = x^2 + 29: Keine reellen Lösungen

Lerne, wie man x - 3 = x^2 + 29 in die Normalform umschreibt, die Diskriminante verwendet und warum die quadratische Gleichung keine reellen Lösungen hat.

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Problem

Solve x3=x2+29x - 3 = x^2 + 29.

Step 1: Move the xx Term

We want all the xx pieces on the same side.

Subtract xx from both sides:

x3x=x2+29xx - 3 - x = x^2 + 29 - x

This gives:

3=x2x+29-3 = x^2 - x + 29

Step 2: Move the Number Term

Now we want one side to be 00.

Add 33 to both sides:

3+3=x2x+29+3-3 + 3 = x^2 - x + 29 + 3

This gives:

0=x2x+320 = x^2 - x + 32

So the standard form is:

x2x+32=0x^2 - x + 32 = 0

Step 3: Name the Three Coefficients

The standard form is:

ax2+bx+c=0ax^2 + bx + c = 0

For:

x2x+32=0x^2 - x + 32 = 0

the three numbers are:

a=1a = 1

b=1b = -1

c=32c = 32

Step 4: Build the Discriminant

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

b24acb^2 - 4ac

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

(1)24(1)(32)(-1)^2 - 4(1)(32)

Now simplify:

1128=1271 - 128 = -127

So the discriminant is:

127-127

Step 5: Read the Answer Type

The discriminant is negative:

127<0-127 < 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=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substitute the values:

x=(1)±1272(1)x = \frac{-(-1) \pm \sqrt{-127}}{2(1)}

Simplify:

x=1±1272x = \frac{1 \pm \sqrt{-127}}{2}

Since 127=i127\sqrt{-127} = i\sqrt{127}, the solutions are:

x=1±i1272x = \frac{1 \pm i\sqrt{127}}{2}

Konzepte

Linear Equations in One Variable

Linear equations where the variable may appear on both sides, inside parentheses, or with unknown coefficients. Solving requires distributing, combining like terms, and isolating the variable. Includes analyzing whether an equation has one solution, no solution, or infinitely many solutions.

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