Question Solve $x^{2}+28x=5$ by completing the square. Select the correct answer: A) $x=14 \pm \sqrt{191}$ B) $x=-14 \pm \sqrt{191}$ C) $x=-14 \pm \sqrt{201}$ D) $x=14 \pm \sqrt{201}$

Studdy Solution

STEP 1

Assumptions
1. We are given the quadratic equation $x^{2} + 28x = 5$ .
2. We need to solve the equation by completing the square.
3. The solution will be in the form $x = a \pm \sqrt{b}$ , where $a$ and $b$ are numbers we need to determine.

STEP 2

To complete the square, the equation must be in the form $x^2 + bx + (b/2)^2 = (b/2)^2 + c$ , where $c$ is a constant. We will add and subtract $(28/2)^2$ on the left side of the equation to complete the square.

STEP 3

Calculate $(28/2)^2$ .
$\left(\frac{28}{2}\right)^2 = 14^2 = 196$

STEP 4

Add and subtract $196$ to the left side of the equation.
$x^{2} + 28x + 196 - 196 = 5$

STEP 5

Rewrite the equation grouping the perfect square trinomial and moving the constants to the right side.
$x^{2} + 28x + 196 = 5 + 196$

STEP 6

Simplify the right side of the equation.
$x^{2} + 28x + 196 = 201$

STEP 7

The left side of the equation is now a perfect square trinomial, which can be factored as $(x + 14)^2$ .
$(x + 14)^2 = 201$

STEP 8

Take the square root of both sides of the equation to solve for $x$ .
$\sqrt{(x + 14)^2} = \pm\sqrt{201}$

STEP 9

Simplify the square root on the left side and keep the $\pm$ on the right side.
$x + 14 = \pm\sqrt{201}$

STEP 10

Subtract $14$ from both sides to isolate $x$ .
$x = -14 \pm\sqrt{201}$
The correct answer is C. $x = -14 \pm\sqrt{201}$ .

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Question Solve $x^{2}+28x=5$ by completing the square. Select the correct answer: A) $x=14 \pm \sqrt{191}$ B) $x=-14 \pm \sqrt{191}$ C) $x=-14 \pm \sqrt{201}$ D) $x=14 \pm \sqrt{201}$

Studdy Solution

STEP 1

Assumptions
1. We are given the quadratic equation $x^{2} + 28x = 5$ .
2. We need to solve the equation by completing the square.
3. The solution will be in the form $x = a \pm \sqrt{b}$ , where $a$ and $b$ are numbers we need to determine.

STEP 2

To complete the square, the equation must be in the form $x^2 + bx + (b/2)^2 = (b/2)^2 + c$ , where $c$ is a constant. We will add and subtract $(28/2)^2$ on the left side of the equation to complete the square.

STEP 3

Calculate $(28/2)^2$ .
$\left(\frac{28}{2}\right)^2 = 14^2 = 196$

STEP 4

Add and subtract $196$ to the left side of the equation.
$x^{2} + 28x + 196 - 196 = 5$

STEP 5

Rewrite the equation grouping the perfect square trinomial and moving the constants to the right side.
$x^{2} + 28x + 196 = 5 + 196$

STEP 6

Simplify the right side of the equation.
$x^{2} + 28x + 196 = 201$

STEP 7

The left side of the equation is now a perfect square trinomial, which can be factored as $(x + 14)^2$ .
$(x + 14)^2 = 201$

STEP 8

Take the square root of both sides of the equation to solve for $x$ .
$\sqrt{(x + 14)^2} = \pm\sqrt{201}$

STEP 9

Simplify the square root on the left side and keep the $\pm$ on the right side.
$x + 14 = \pm\sqrt{201}$

STEP 10

Subtract $14$ from both sides to isolate $x$ .
$x = -14 \pm\sqrt{201}$
The correct answer is C. $x = -14 \pm\sqrt{201}$ .

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Question Solve x2+28x=5x^{2}+28x=5x2+28x=5 by completing the square. Select the correct answer: A) x=14±191x=14 \pm \sqrt{191}x=14±191​ B) x=−14±191x=-14 \pm \sqrt{191}x=−14±191​ C) x=−14±201x=-14 \pm \sqrt{201}x=−14±201​ D) x=14±201x=14 \pm \sqrt{201}x=14±201​

Studdy Solution

STEP 1

Assumptions1. We are given the quadratic equation x2+28x=5x^{2} + 28x = 5x2+28x=5.2. We need to solve the equation by completing the square.3. The solution will be in the form x=a±bx = a \pm \sqrt{b}x=a±b​, where aaa and bbb are numbers we need to determine.

STEP 2

To complete the square, the equation must be in the form x2+bx+(b/2)2=(b/2)2+cx^2 + bx + (b/2)^2 = (b/2)^2 + cx2+bx+(b/2)2=(b/2)2+c, where ccc is a constant. We will add and subtract (28/2)2(28/2)^2(28/2)2 on the left side of the equation to complete the square.

STEP 3

Calculate (28/2)2(28/2)^2(28/2)2.(282)2=142=196\left(\frac{28}{2}\right)^2 = 14^2 = 196(228​)2=142=196

STEP 4

Add and subtract 196196196 to the left side of the equation.x2+28x+196−196=5x^{2} + 28x + 196 - 196 = 5x2+28x+196−196=5

STEP 5

Rewrite the equation grouping the perfect square trinomial and moving the constants to the right side.x2+28x+196=5+196x^{2} + 28x + 196 = 5 + 196x2+28x+196=5+196

STEP 6

Simplify the right side of the equation.x2+28x+196=201x^{2} + 28x + 196 = 201x2+28x+196=201

STEP 7

The left side of the equation is now a perfect square trinomial, which can be factored as (x+14)2(x + 14)^2(x+14)2.(x+14)2=201(x + 14)^2 = 201(x+14)2=201

STEP 8

Take the square root of both sides of the equation to solve for xxx.(x+14)2=±201\sqrt{(x + 14)^2} = \pm\sqrt{201}(x+14)2​=±201​

STEP 9

Simplify the square root on the left side and keep the ±\pm± on the right side.x+14=±201x + 14 = \pm\sqrt{201}x+14=±201​

STEP 10

Subtract 141414 from both sides to isolate xxx.x=−14±201x = -14 \pm\sqrt{201}x=−14±201​The correct answer is C. x=−14±201x = -14 \pm\sqrt{201}x=−14±201​.

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Question Solve $x^{2}+28x=5$ by completing the square. Select the correct answer: A) $x=14 \pm \sqrt{191}$ B) $x=-14 \pm \sqrt{191}$ C) $x=-14 \pm \sqrt{201}$ D) $x=14 \pm \sqrt{201}$

Assumptions
1. We are given the quadratic equation $x^{2} + 28x = 5$ .
2. We need to solve the equation by completing the square.
3. The solution will be in the form $x = a \pm \sqrt{b}$ , where $a$ and $b$ are numbers we need to determine.

To complete the square, the equation must be in the form $x^2 + bx + (b/2)^2 = (b/2)^2 + c$ , where $c$ is a constant. We will add and subtract $(28/2)^2$ on the left side of the equation to complete the square.

Calculate $(28/2)^2$ .
$\left(\frac{28}{2}\right)^2 = 14^2 = 196$

Add and subtract $196$ to the left side of the equation.
$x^{2} + 28x + 196 - 196 = 5$

Rewrite the equation grouping the perfect square trinomial and moving the constants to the right side.
$x^{2} + 28x + 196 = 5 + 196$

Simplify the right side of the equation.
$x^{2} + 28x + 196 = 201$

The left side of the equation is now a perfect square trinomial, which can be factored as $(x + 14)^2$ .
$(x + 14)^2 = 201$

Take the square root of both sides of the equation to solve for $x$ .
$\sqrt{(x + 14)^2} = \pm\sqrt{201}$

Simplify the square root on the left side and keep the $\pm$ on the right side.
$x + 14 = \pm\sqrt{201}$

Subtract $14$ from both sides to isolate $x$ .
$x = -14 \pm\sqrt{201}$
The correct answer is C. $x = -14 \pm\sqrt{201}$ .