QuestionProve that the solutions of $z^{2}-2 z+10=0$ are complex (not real).

Studdy Solution

STEP 1

Assumptions1. The equation is $z^{}- z+10=0$ . We need to prove that the roots of this equation are not real.

STEP 2

The roots of a quadratic equation $ax^{2}+bx+c=0$ can be found using the quadratic formula $z = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

STEP 3

In our equation, $a=1$ , $b=-2$ , and $c=10$ . Let's substitute these values into the quadratic formula.
$z = \frac{-(-2) \pm \sqrt{(-2)^{2}-(1)(10)}}{2(1)}$

STEP 4

implify the equation.
$z = \frac{2 \pm \sqrt{4-40}}{2}$

STEP 5

Further simplify the equation.
$z = \frac{2 \pm \sqrt{-36}}{2}$

STEP 6

The square root of a negative number is not a real number. It's an imaginary number. Therefore, the roots of the equation $z^{2}-2 z+10=0$ are not real.

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QuestionProve that the solutions of $z^{2}-2 z+10=0$ are complex (not real).

Studdy Solution

STEP 1

Assumptions1. The equation is $z^{}- z+10=0$ . We need to prove that the roots of this equation are not real.

STEP 2

The roots of a quadratic equation $ax^{2}+bx+c=0$ can be found using the quadratic formula $z = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

STEP 3

In our equation, $a=1$ , $b=-2$ , and $c=10$ . Let's substitute these values into the quadratic formula.
$z = \frac{-(-2) \pm \sqrt{(-2)^{2}-(1)(10)}}{2(1)}$

STEP 4

implify the equation.
$z = \frac{2 \pm \sqrt{4-40}}{2}$

STEP 5

Further simplify the equation.
$z = \frac{2 \pm \sqrt{-36}}{2}$

STEP 6

The square root of a negative number is not a real number. It's an imaginary number. Therefore, the roots of the equation $z^{2}-2 z+10=0$ are not real.

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QuestionProve that the solutions of z2−2z+10=0z^{2}-2 z+10=0z2−2z+10=0 are complex (not real).

Studdy Solution

STEP 1

Assumptions1. The equation is z−z+10=0z^{}- z+10=0z−z+10=0 . We need to prove that the roots of this equation are not real.

STEP 2

The roots of a quadratic equation ax2+bx+c=0ax^{2}+bx+c=0ax2+bx+c=0 can be found using the quadratic formulaz=−b±b2−4ac2az = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}z=2a−b±b2−4ac​​

STEP 3

In our equation, a=1a=1a=1, b=−2b=-2b=−2, and c=10c=10c=10. Let's substitute these values into the quadratic formula.z=−(−2)±(−2)2−(1)(10)2(1)z = \frac{-(-2) \pm \sqrt{(-2)^{2}-(1)(10)}}{2(1)}z=2(1)−(−2)±(−2)2−(1)(10)​​

STEP 4

implify the equation.z=2±4−402z = \frac{2 \pm \sqrt{4-40}}{2}z=22±4−40​​

STEP 5

Further simplify the equation.z=2±−362z = \frac{2 \pm \sqrt{-36}}{2}z=22±−36​​

STEP 6

The square root of a negative number is not a real number. It's an imaginary number. Therefore, the roots of the equation z2−2z+10=0z^{2}-2 z+10=0z2−2z+10=0 are not real.

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QuestionProve that the solutions of $z^{2}-2 z+10=0$ are complex (not real).

Assumptions1. The equation is $z^{}- z+10=0$ . We need to prove that the roots of this equation are not real.

The roots of a quadratic equation $ax^{2}+bx+c=0$ can be found using the quadratic formula $z = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

In our equation, $a=1$ , $b=-2$ , and $c=10$ . Let's substitute these values into the quadratic formula.
$z = \frac{-(-2) \pm \sqrt{(-2)^{2}-(1)(10)}}{2(1)}$

implify the equation.
$z = \frac{2 \pm \sqrt{4-40}}{2}$

Further simplify the equation.
$z = \frac{2 \pm \sqrt{-36}}{2}$

The square root of a negative number is not a real number. It's an imaginary number. Therefore, the roots of the equation $z^{2}-2 z+10=0$ are not real.