QuestionSolve the equation $x^{2}-12 x-14=0$ and choose the correct solution set from the options provided.

Studdy Solution

STEP 1

Assumptions1. The equation is a quadratic equation of the form $ax^ + bx + c =0$ . The coefficients are a =1, b = -12, and c = -143. The roots of the equation can be found using the quadratic formula $x = \frac{-b \pm \sqrt{b^ -4ac}}{a}$

STEP 2

Let's start by plugging the coefficients into the quadratic formula.
$x = \frac{-(-12) \pm \sqrt{(-12)^2 -41(-14)}}{2*1}$

STEP 3

implify the equation.
$x = \frac{12 \pm \sqrt{144 +56}}{2}$

STEP 4

Add the numbers under the square root.
$x = \frac{12 \pm \sqrt{200}}{2}$

STEP 5

implify the square root. Note that $200 =450 =4510 =4525 =2^25^22$ .
$x = \frac{12 \pm25\sqrt{2}}{2}$

STEP 6

implify the equation by dividing every term by2.
$x =6 \pm5\sqrt{2}$ The roots of the equation are $6 +5\sqrt{2}$ and $6 -5\sqrt{2}$ , which corresponds to option d.

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QuestionSolve the equation $x^{2}-12 x-14=0$ and choose the correct solution set from the options provided.

Studdy Solution

STEP 1

Assumptions1. The equation is a quadratic equation of the form $ax^ + bx + c =0$ . The coefficients are a =1, b = -12, and c = -143. The roots of the equation can be found using the quadratic formula $x = \frac{-b \pm \sqrt{b^ -4ac}}{a}$

STEP 2

Let's start by plugging the coefficients into the quadratic formula.
$x = \frac{-(-12) \pm \sqrt{(-12)^2 -41(-14)}}{2*1}$

STEP 3

implify the equation.
$x = \frac{12 \pm \sqrt{144 +56}}{2}$

STEP 4

Add the numbers under the square root.
$x = \frac{12 \pm \sqrt{200}}{2}$

STEP 5

implify the square root. Note that $200 =450 =4510 =4525 =2^25^22$ .
$x = \frac{12 \pm25\sqrt{2}}{2}$

STEP 6

implify the equation by dividing every term by2.
$x =6 \pm5\sqrt{2}$ The roots of the equation are $6 +5\sqrt{2}$ and $6 -5\sqrt{2}$ , which corresponds to option d.

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QuestionSolve the equation x2−12x−14=0x^{2}-12 x-14=0x2−12x−14=0 and choose the correct solution set from the options provided.

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 a =1, b = -12, and c = -143. The roots of the equation can be found using the quadratic formula x=−b±b−4acax = \frac{-b \pm \sqrt{b^ -4ac}}{a}x=a−b±b−4ac​​

STEP 2

Let's start by plugging the coefficients into the quadratic formula.x=−(−12)±(−12)2−4∗1∗(−14)2∗1x = \frac{-(-12) \pm \sqrt{(-12)^2 -4*1*(-14)}}{2*1}x=2∗1−(−12)±(−12)2−4∗1∗(−14)​​

STEP 3

implify the equation.x=12±144+562x = \frac{12 \pm \sqrt{144 +56}}{2}x=212±144+56​​

STEP 4

Add the numbers under the square root.x=12±2002x = \frac{12 \pm \sqrt{200}}{2}x=212±200​​

STEP 5

implify the square root. Note that 200=4∗50=4∗5∗10=4∗5∗2∗5=22∗52∗2200 =4*50 =4*5*10 =4*5*2*5 =2^2*5^2*2200=4∗50=4∗5∗10=4∗5∗2∗5=22∗52∗2.x=12±2∗5∗22x = \frac{12 \pm2*5*\sqrt{2}}{2}x=212±2∗5∗2​​

STEP 6

implify the equation by dividing every term by2.x=6±52x =6 \pm5\sqrt{2}x=6±52​The roots of the equation are 6+526 +5\sqrt{2}6+52​ and 6−526 -5\sqrt{2}6−52​, which corresponds to option d.

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QuestionSolve the equation $x^{2}-12 x-14=0$ and choose the correct solution set from the options provided.

Assumptions1. The equation is a quadratic equation of the form $ax^ + bx + c =0$ . The coefficients are a =1, b = -12, and c = -143. The roots of the equation can be found using the quadratic formula $x = \frac{-b \pm \sqrt{b^ -4ac}}{a}$

Let's start by plugging the coefficients into the quadratic formula.
$x = \frac{-(-12) \pm \sqrt{(-12)^2 -41(-14)}}{2*1}$

implify the equation.
$x = \frac{12 \pm \sqrt{144 +56}}{2}$

Add the numbers under the square root.
$x = \frac{12 \pm \sqrt{200}}{2}$

implify the square root. Note that $200 =450 =4510 =4525 =2^25^22$ .
$x = \frac{12 \pm25\sqrt{2}}{2}$

implify the equation by dividing every term by2.
$x =6 \pm5\sqrt{2}$ The roots of the equation are $6 +5\sqrt{2}$ and $6 -5\sqrt{2}$ , which corresponds to option d.