Solved on Sep 16, 2023

Solve the quadratic equation $15 x^{2}=11 x-2$ and find the fractions $2 / 5, 1 / 3, -2 / 5, -1 / 3$ .

STEP 1

Assumptions1. The equation given is $15x^{}=11x-$ . We are looking for the roots of the equation, which are the values of $x$ that satisfy the equation.

STEP 2

To solve for $x$ , we need to rearrange the equation into a standard quadratic form $ax^{2} + bx + c =0$ . $15x^{2} -11x +2 =0$

STEP 3

The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2} -ac}}{2a}$ where $a$ , $b$ , and $c$ are coefficients of the quadratic equation. In our case, $a =15$ , $b = -11$ , and $c =2$ .

STEP 4

First, we calculate the value under the square root in the quadratic formula, which is $b^{2} -4ac$ .
$b^{2} -4ac = (-11)^{2} -4 \cdot15 \cdot2$

STEP 5

Calculate the value of $b^{2} -4ac$ .
$b^{2} -4ac =121 -120 =1$

STEP 6

Now, we can substitute $a$ , $b$ , and $b^{2} -4ac$ into the quadratic formula to solve for $x$ .
$x = \frac{-(-11) \pm \sqrt{1}}{2 \cdot15}$

STEP 7

implify the equation.
$x = \frac{11 \pm1}{30}$

STEP 8

The $\pm$ sign means there are two solutions one with the plus sign and one with the minus sign. Let's calculate both.
For the plus sign $x = \frac{11 +1}{30} = \frac{12}{30} = \frac{2}{5}$ For the minus sign $x = \frac{11 -1}{30} = \frac{10}{30} = \frac{1}{3}$ So, the solutions to the equation $15x^{2}=11x-2$ are $x = \frac{2}{5}$ and $x = \frac{1}{3}$ .

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Solve the quadratic equation $15 x^{2}=11 x-2$ and find the fractions $2 / 5, 1 / 3, -2 / 5, -1 / 3$ .

STEP 1

Assumptions1. The equation given is $15x^{}=11x-$ . We are looking for the roots of the equation, which are the values of $x$ that satisfy the equation.

STEP 2

To solve for $x$ , we need to rearrange the equation into a standard quadratic form $ax^{2} + bx + c =0$ . $15x^{2} -11x +2 =0$

STEP 3

The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2} -ac}}{2a}$ where $a$ , $b$ , and $c$ are coefficients of the quadratic equation. In our case, $a =15$ , $b = -11$ , and $c =2$ .

STEP 4

First, we calculate the value under the square root in the quadratic formula, which is $b^{2} -4ac$ .
$b^{2} -4ac = (-11)^{2} -4 \cdot15 \cdot2$

STEP 5

Calculate the value of $b^{2} -4ac$ .
$b^{2} -4ac =121 -120 =1$

STEP 6

Now, we can substitute $a$ , $b$ , and $b^{2} -4ac$ into the quadratic formula to solve for $x$ .
$x = \frac{-(-11) \pm \sqrt{1}}{2 \cdot15}$

STEP 7

implify the equation.
$x = \frac{11 \pm1}{30}$

STEP 8

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Solve the quadratic equation 15x2=11x−215 x^{2}=11 x-215x2=11x−2 and find the fractions 2/5,1/3,−2/5,−1/32 / 5, 1 / 3, -2 / 5, -1 / 32/5,1/3,−2/5,−1/3.

STEP 1

Assumptions1. The equation given is 15x=11x−15x^{}=11x-15x=11x− . We are looking for the roots of the equation, which are the values of xxx that satisfy the equation.

STEP 2

To solve for xxx, we need to rearrange the equation into a standard quadratic form ax2+bx+c=0ax^{2} + bx + c =0ax2+bx+c=0.15x2−11x+2=015x^{2} -11x +2 =015x2−11x+2=0

STEP 3

The quadratic formula is given byx=−b±b2−ac2ax = \frac{-b \pm \sqrt{b^{2} -ac}}{2a}x=2a−b±b2−ac​​where aaa, bbb, and ccc are coefficients of the quadratic equation. In our case, a=15a =15a=15, b=−11b = -11b=−11, and c=2c =2c=2.

STEP 4

First, we calculate the value under the square root in the quadratic formula, which is b2−4acb^{2} -4acb2−4ac.b2−4ac=(−11)2−4⋅15⋅2b^{2} -4ac = (-11)^{2} -4 \cdot15 \cdot2b2−4ac=(−11)2−4⋅15⋅2

STEP 5

Calculate the value of b2−4acb^{2} -4acb2−4ac.b2−4ac=121−120=1b^{2} -4ac =121 -120 =1b2−4ac=121−120=1

STEP 6

Now, we can substitute aaa, bbb, and b2−4acb^{2} -4acb2−4ac into the quadratic formula to solve for xxx.x=−(−11)±12⋅15x = \frac{-(-11) \pm \sqrt{1}}{2 \cdot15}x=2⋅15−(−11)±1​​

STEP 7

implify the equation.x=11±130x = \frac{11 \pm1}{30}x=3011±1​

STEP 8

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Solve the quadratic equation $15 x^{2}=11 x-2$ and find the fractions $2 / 5, 1 / 3, -2 / 5, -1 / 3$ .

Assumptions1. The equation given is $15x^{}=11x-$ . We are looking for the roots of the equation, which are the values of $x$ that satisfy the equation.

To solve for $x$ , we need to rearrange the equation into a standard quadratic form $ax^{2} + bx + c =0$ . $15x^{2} -11x +2 =0$

The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2} -ac}}{2a}$ where $a$ , $b$ , and $c$ are coefficients of the quadratic equation. In our case, $a =15$ , $b = -11$ , and $c =2$ .

First, we calculate the value under the square root in the quadratic formula, which is $b^{2} -4ac$ .
$b^{2} -4ac = (-11)^{2} -4 \cdot15 \cdot2$

Calculate the value of $b^{2} -4ac$ .
$b^{2} -4ac =121 -120 =1$

Now, we can substitute $a$ , $b$ , and $b^{2} -4ac$ into the quadratic formula to solve for $x$ .
$x = \frac{-(-11) \pm \sqrt{1}}{2 \cdot15}$

implify the equation.
$x = \frac{11 \pm1}{30}$