QuestionSolve the equation $x^{2}-x+4=0$ .

Studdy Solution

STEP 1

Assumptions1. The equation is a quadratic equation of the form $ax^{} + bx + c =0$ . The coefficients are real numbers $a=1$ , $b=-1$ , and $c=4$
3. The solutions for $x$ can be real or complex numbers

STEP 2

The solutions to a quadratic equation are given by the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}$

STEP 3

Now, plug in the given values for $a$ , $b$ , and $c$ into the quadratic formula.
$x = \frac{-(-1) \pm \sqrt{(-1)^{2} -(1)()}}{2(1)}$

STEP 4

implify the equation.
$x = \frac{1 \pm \sqrt{1 -16}}{2}$

STEP 5

Calculate the value under the square root.
$x = \frac{1 \pm \sqrt{-15}}{2}$

STEP 6

Since the value under the square root is negative, the solutions will be complex. We can express the square root of a negative number as $i$ times the square root of the absolute value of that number.
$x = \frac{1 \pm i\sqrt{15}}{2}$

STEP 7

implify the equation to find the solutions for $x$ .
$x = \frac{1}{2} \pm \frac{i\sqrt{15}}{2}$ The solutions for the equation $x^{2}-x+4=0$ are $x = \frac{1}{2} + \frac{i\sqrt{15}}{2}$ and $x = \frac{1}{2} - \frac{i\sqrt{15}}{2}$ .

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QuestionSolve the equation $x^{2}-x+4=0$ .

Studdy Solution

STEP 1

Assumptions1. The equation is a quadratic equation of the form $ax^{} + bx + c =0$ . The coefficients are real numbers $a=1$ , $b=-1$ , and $c=4$
3. The solutions for $x$ can be real or complex numbers

STEP 2

The solutions to a quadratic equation are given by the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}$

STEP 3

Now, plug in the given values for $a$ , $b$ , and $c$ into the quadratic formula.
$x = \frac{-(-1) \pm \sqrt{(-1)^{2} -(1)()}}{2(1)}$

STEP 4

implify the equation.
$x = \frac{1 \pm \sqrt{1 -16}}{2}$

STEP 5

Calculate the value under the square root.
$x = \frac{1 \pm \sqrt{-15}}{2}$

STEP 6

Since the value under the square root is negative, the solutions will be complex. We can express the square root of a negative number as $i$ times the square root of the absolute value of that number.
$x = \frac{1 \pm i\sqrt{15}}{2}$

STEP 7

implify the equation to find the solutions for $x$ .
$x = \frac{1}{2} \pm \frac{i\sqrt{15}}{2}$ The solutions for the equation $x^{2}-x+4=0$ are $x = \frac{1}{2} + \frac{i\sqrt{15}}{2}$ and $x = \frac{1}{2} - \frac{i\sqrt{15}}{2}$ .

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QuestionSolve the equation x2−x+4=0x^{2}-x+4=0x2−x+4=0.

Studdy Solution

STEP 1

Assumptions1. The equation is a quadratic equation of the form ax+bx+c=0ax^{} + bx + c =0ax+bx+c=0 . The coefficients are real numbers a=1a=1a=1, b=−1b=-1b=−1, and c=4c=4c=43. The solutions for xxx can be real or complex numbers

STEP 2

The solutions to a quadratic equation are given by the quadratic formulax=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}x=2a−b±b2−4ac​​

STEP 3

Now, plug in the given values for aaa, bbb, and ccc into the quadratic formula.x=−(−1)±(−1)2−(1)()2(1)x = \frac{-(-1) \pm \sqrt{(-1)^{2} -(1)()}}{2(1)}x=2(1)−(−1)±(−1)2−(1)()​​

STEP 4

implify the equation.x=1±1−162x = \frac{1 \pm \sqrt{1 -16}}{2}x=21±1−16​​

STEP 5

Calculate the value under the square root.x=1±−152x = \frac{1 \pm \sqrt{-15}}{2}x=21±−15​​

STEP 6

Since the value under the square root is negative, the solutions will be complex. We can express the square root of a negative number as iii times the square root of the absolute value of that number.x=1±i152x = \frac{1 \pm i\sqrt{15}}{2}x=21±i15​​

STEP 7

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QuestionSolve the equation $x^{2}-x+4=0$ .

Assumptions1. The equation is a quadratic equation of the form $ax^{} + bx + c =0$ . The coefficients are real numbers $a=1$ , $b=-1$ , and $c=4$
3. The solutions for $x$ can be real or complex numbers

The solutions to a quadratic equation are given by the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}$

Now, plug in the given values for $a$ , $b$ , and $c$ into the quadratic formula.
$x = \frac{-(-1) \pm \sqrt{(-1)^{2} -(1)()}}{2(1)}$

implify the equation.
$x = \frac{1 \pm \sqrt{1 -16}}{2}$

Calculate the value under the square root.
$x = \frac{1 \pm \sqrt{-15}}{2}$

Since the value under the square root is negative, the solutions will be complex. We can express the square root of a negative number as $i$ times the square root of the absolute value of that number.
$x = \frac{1 \pm i\sqrt{15}}{2}$