Solved on Sep 16, 2023

Solve for $x$ in the quadratic equation $2 x^{2} - 5 x + 1 = 3$ .

STEP 1

Assumptions1. The equation is quadratic in form, i.e., $ax^ + bx + c =0$ . We are solving for $x$
3. The coefficients of the equation are real numbers

STEP 2

First, we need to rearrange the equation into the standard form of a quadratic equation $ax^2 + bx + c =0$ . $2x^2 -5x +1 - =0$

STEP 3

implify the equation.
$2x^2 -5x -2 =0$

STEP 4

The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation.

STEP 5

Identify the coefficients in the equation $2x^2 -5x -2 =0$ . $a =2$ , $b = -5$ , and $c = -2$

STEP 6

Substitute the coefficients into the quadratic formula.
$x = \frac{-(-5) \pm \sqrt{(-5)^2 -42(-2)}}{2*2}$

STEP 7

implify the equation.
$x = \frac{5 \pm \sqrt{25 +16}}{4}$

STEP 8

Further simplify the equation.
$x = \frac{5 \pm \sqrt{41}}{4}$

STEP 9

This gives the two solutions for $x$ .
$x = \frac{5 + \sqrt{41}}{4} \quad or \quad x = \frac{5 - \sqrt{41}}{4}$ So, the solutions to the equation $2x^2 -5x + =3$ are $x = \frac{5 + \sqrt{41}}{4}$ and $x = \frac{5 - \sqrt{41}}{4}$ .

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Solve for $x$ in the quadratic equation $2 x^{2} - 5 x + 1 = 3$ .

STEP 1

Assumptions1. The equation is quadratic in form, i.e., $ax^ + bx + c =0$ . We are solving for $x$
3. The coefficients of the equation are real numbers

STEP 2

First, we need to rearrange the equation into the standard form of a quadratic equation $ax^2 + bx + c =0$ . $2x^2 -5x +1 - =0$

STEP 3

implify the equation.
$2x^2 -5x -2 =0$

STEP 4

The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation.

STEP 5

Identify the coefficients in the equation $2x^2 -5x -2 =0$ . $a =2$ , $b = -5$ , and $c = -2$

STEP 6

Substitute the coefficients into the quadratic formula.
$x = \frac{-(-5) \pm \sqrt{(-5)^2 -42(-2)}}{2*2}$

STEP 7

implify the equation.
$x = \frac{5 \pm \sqrt{25 +16}}{4}$

STEP 8

Further simplify the equation.
$x = \frac{5 \pm \sqrt{41}}{4}$

STEP 9

This gives the two solutions for $x$ .
$x = \frac{5 + \sqrt{41}}{4} \quad or \quad x = \frac{5 - \sqrt{41}}{4}$ So, the solutions to the equation $2x^2 -5x + =3$ are $x = \frac{5 + \sqrt{41}}{4}$ and $x = \frac{5 - \sqrt{41}}{4}$ .

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Solve for xxx in the quadratic equation 2x2−5x+1=32 x^{2} - 5 x + 1 = 32x2−5x+1=3.

STEP 1

Assumptions1. The equation is quadratic in form, i.e., ax+bx+c=0ax^ + bx + c =0ax+bx+c=0 . We are solving for xxx3. The coefficients of the equation are real numbers

STEP 2

First, we need to rearrange the equation into the standard form of a quadratic equation ax2+bx+c=0ax^2 + bx + c =0ax2+bx+c=0.2x2−5x+1−=02x^2 -5x +1 - =02x2−5x+1−=0

STEP 3

implify the equation.2x2−5x−2=02x^2 -5x -2 =02x2−5x−2=0

STEP 4

The quadratic formula is given byx=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}x=2a−b±b2−4ac​​where aaa, bbb, and ccc are the coefficients of the quadratic equation.

STEP 5

Identify the coefficients in the equation 2x2−5x−2=02x^2 -5x -2 =02x2−5x−2=0.a=2a =2a=2, b=−5b = -5b=−5, and c=−2c = -2c=−2

STEP 6

Substitute the coefficients into the quadratic formula.x=−(−5)±(−5)2−4∗2∗(−2)2∗2x = \frac{-(-5) \pm \sqrt{(-5)^2 -4*2*(-2)}}{2*2}x=2∗2−(−5)±(−5)2−4∗2∗(−2)​​

STEP 7

implify the equation.x=5±25+164x = \frac{5 \pm \sqrt{25 +16}}{4}x=45±25+16​​

STEP 8

Further simplify the equation.x=5±414x = \frac{5 \pm \sqrt{41}}{4}x=45±41​​

STEP 9

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Solve for $x$ in the quadratic equation $2 x^{2} - 5 x + 1 = 3$ .

Assumptions1. The equation is quadratic in form, i.e., $ax^ + bx + c =0$ . We are solving for $x$
3. The coefficients of the equation are real numbers

First, we need to rearrange the equation into the standard form of a quadratic equation $ax^2 + bx + c =0$ . $2x^2 -5x +1 - =0$

implify the equation.
$2x^2 -5x -2 =0$

The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation.

Identify the coefficients in the equation $2x^2 -5x -2 =0$ . $a =2$ , $b = -5$ , and $c = -2$

Substitute the coefficients into the quadratic formula.
$x = \frac{-(-5) \pm \sqrt{(-5)^2 -42(-2)}}{2*2}$

implify the equation.
$x = \frac{5 \pm \sqrt{25 +16}}{4}$

Further simplify the equation.
$x = \frac{5 \pm \sqrt{41}}{4}$