QuestionSolve for $x$ in the equation $x^{2}-14 x+49=0$ .

Studdy Solution

STEP 1

Assumptions1. The equation is $x^{}-14 x+49=0$

STEP 2

This equation is a quadratic equation in the form of $ax^2+bx+c=0$ . In this case, $a=1$ , $b=-14$ , and $c=49$ .

STEP 3

The quadratic formula is used to solve for $x$ in a quadratic equation. The formula is $x = \frac{-b \pm \sqrt{b^2 -ac}}{2a}$

STEP 4

Substitute $a=1$ , $b=-14$ , and $c=49$ into the quadratic formula.
$x = \frac{-(-14) \pm \sqrt{(-14)^2 -4149}}{2*1}$

STEP 5

implify the equation.
$x = \frac{14 \pm \sqrt{196 -196}}{2}$

STEP 6

Calculate the value under the square root.
$x = \frac{14 \pm \sqrt{0}}{2}$

STEP 7

implify the square root.
$x = \frac{14 \pm0}{2}$

STEP 8

Calculate the two possible values for $x$ .
$x = \frac{14 +0}{2} =7$ $x = \frac{14 -0}{2} =7$ In this case, both solutions are the same. Therefore, the solution to the equation $x^{2}-14 x+49=0$ is $x=7$ .

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QuestionSolve for $x$ in the equation $x^{2}-14 x+49=0$ .

Studdy Solution

STEP 1

Assumptions1. The equation is $x^{}-14 x+49=0$

STEP 2

This equation is a quadratic equation in the form of $ax^2+bx+c=0$ . In this case, $a=1$ , $b=-14$ , and $c=49$ .

STEP 3

The quadratic formula is used to solve for $x$ in a quadratic equation. The formula is $x = \frac{-b \pm \sqrt{b^2 -ac}}{2a}$

STEP 4

Substitute $a=1$ , $b=-14$ , and $c=49$ into the quadratic formula.
$x = \frac{-(-14) \pm \sqrt{(-14)^2 -4149}}{2*1}$

STEP 5

implify the equation.
$x = \frac{14 \pm \sqrt{196 -196}}{2}$

STEP 6

Calculate the value under the square root.
$x = \frac{14 \pm \sqrt{0}}{2}$

STEP 7

implify the square root.
$x = \frac{14 \pm0}{2}$

STEP 8

Calculate the two possible values for $x$ .
$x = \frac{14 +0}{2} =7$ $x = \frac{14 -0}{2} =7$ In this case, both solutions are the same. Therefore, the solution to the equation $x^{2}-14 x+49=0$ is $x=7$ .

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QuestionSolve for xxx in the equation x2−14x+49=0x^{2}-14 x+49=0x2−14x+49=0.

Studdy Solution

STEP 1

Assumptions1. The equation is x−14x+49=0x^{}-14 x+49=0x−14x+49=0

STEP 2

This equation is a quadratic equation in the form of ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0. In this case, a=1a=1a=1, b=−14b=-14b=−14, and c=49c=49c=49.

STEP 3

The quadratic formula is used to solve for xxx in a quadratic equation. The formula isx=−b±b2−ac2ax = \frac{-b \pm \sqrt{b^2 -ac}}{2a}x=2a−b±b2−ac​​

STEP 4

Substitute a=1a=1a=1, b=−14b=-14b=−14, and c=49c=49c=49 into the quadratic formula.x=−(−14)±(−14)2−4∗1∗492∗1x = \frac{-(-14) \pm \sqrt{(-14)^2 -4*1*49}}{2*1}x=2∗1−(−14)±(−14)2−4∗1∗49​​

STEP 5

implify the equation.x=14±196−1962x = \frac{14 \pm \sqrt{196 -196}}{2}x=214±196−196​​

STEP 6

Calculate the value under the square root.x=14±02x = \frac{14 \pm \sqrt{0}}{2}x=214±0​​

STEP 7

implify the square root.x=14±02x = \frac{14 \pm0}{2}x=214±0​

STEP 8

Calculate the two possible values for xxx.x=14+02=7x = \frac{14 +0}{2} =7x=214+0​=7x=14−02=7x = \frac{14 -0}{2} =7x=214−0​=7In this case, both solutions are the same. Therefore, the solution to the equation x2−14x+49=0x^{2}-14 x+49=0x2−14x+49=0 is x=7x=7x=7.

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QuestionSolve for $x$ in the equation $x^{2}-14 x+49=0$ .

Assumptions1. The equation is $x^{}-14 x+49=0$

This equation is a quadratic equation in the form of $ax^2+bx+c=0$ . In this case, $a=1$ , $b=-14$ , and $c=49$ .

The quadratic formula is used to solve for $x$ in a quadratic equation. The formula is $x = \frac{-b \pm \sqrt{b^2 -ac}}{2a}$

Substitute $a=1$ , $b=-14$ , and $c=49$ into the quadratic formula.
$x = \frac{-(-14) \pm \sqrt{(-14)^2 -4149}}{2*1}$

implify the equation.
$x = \frac{14 \pm \sqrt{196 -196}}{2}$

Calculate the value under the square root.
$x = \frac{14 \pm \sqrt{0}}{2}$

implify the square root.
$x = \frac{14 \pm0}{2}$

Calculate the two possible values for $x$ .
$x = \frac{14 +0}{2} =7$ $x = \frac{14 -0}{2} =7$ In this case, both solutions are the same. Therefore, the solution to the equation $x^{2}-14 x+49=0$ is $x=7$ .