Solved on Oct 30, 2023

Solve for $u$ where $2 u^{2}+14 u+24=(u+3)^{2}$ . If multiple solutions, list them separated by commas.

STEP 1

Assumptions1. The equation is quadratic in nature. We are looking for real solutions for $u$

STEP 2

We are given the equation $2 u^{2}+14 u+24=(u+)^{2}$

STEP 3

First, we need to expand the right side of the equation $(u+3)^{2}$ .
$(u+3)^{2} = u^{2} +23u +3^{2}$

STEP 4

Calculate the expanded form of $(u+3)^{2}$ .
$(u+3)^{2} = u^{2} +6u +9$

STEP 5

Now, substitute the expanded form of $(u+3)^{2}$ back into the original equation.
$2 u^{2}+14 u+24 = u^{2} +u +9$

STEP 6

To solve for $u$ , we need to bring all terms to one side of the equation. Subtract $u^{2}$ , $6u$ , and $9$ from both sides.
$2 u^{2} - u^{2} +14 u -6u +24 -9 =0$

STEP 7

implify the equation.
$u^{2} +u +15 =0$

STEP 8

Now, we need to factor the quadratic equation. We are looking for two numbers that multiply to $15$ and add to $8$ .
$(u +5)(u +3) =0$

STEP 9

Set each factor equal to zero and solve for $u$ .
$u +5 =$ $u +3 =$

STEP 10

olve for $u$ in each equation.
$u = -5$ $u = -3$ The solutions for $u$ are $-5$ and $-3$ .

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Solve for $u$ where $2 u^{2}+14 u+24=(u+3)^{2}$ . If multiple solutions, list them separated by commas.

STEP 1

Assumptions1. The equation is quadratic in nature. We are looking for real solutions for $u$

STEP 2

We are given the equation $2 u^{2}+14 u+24=(u+)^{2}$

STEP 3

First, we need to expand the right side of the equation $(u+3)^{2}$ .
$(u+3)^{2} = u^{2} +23u +3^{2}$

STEP 4

Calculate the expanded form of $(u+3)^{2}$ .
$(u+3)^{2} = u^{2} +6u +9$

STEP 5

Now, substitute the expanded form of $(u+3)^{2}$ back into the original equation.
$2 u^{2}+14 u+24 = u^{2} +u +9$

STEP 6

To solve for $u$ , we need to bring all terms to one side of the equation. Subtract $u^{2}$ , $6u$ , and $9$ from both sides.
$2 u^{2} - u^{2} +14 u -6u +24 -9 =0$

STEP 7

implify the equation.
$u^{2} +u +15 =0$

STEP 8

Now, we need to factor the quadratic equation. We are looking for two numbers that multiply to $15$ and add to $8$ .
$(u +5)(u +3) =0$

STEP 9

Set each factor equal to zero and solve for $u$ .
$u +5 =$ $u +3 =$

STEP 10

olve for $u$ in each equation.
$u = -5$ $u = -3$ The solutions for $u$ are $-5$ and $-3$ .

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Solve for uuu where 2u2+14u+24=(u+3)22 u^{2}+14 u+24=(u+3)^{2}2u2+14u+24=(u+3)2. If multiple solutions, list them separated by commas.

STEP 1

Assumptions1. The equation is quadratic in nature. We are looking for real solutions for uuu

STEP 2

We are given the equation2u2+14u+24=(u+)22 u^{2}+14 u+24=(u+)^{2}2u2+14u+24=(u+)2

STEP 3

First, we need to expand the right side of the equation (u+3)2(u+3)^{2}(u+3)2.(u+3)2=u2+2∗3∗u+32(u+3)^{2} = u^{2} +2*3*u +3^{2}(u+3)2=u2+2∗3∗u+32

STEP 4

Calculate the expanded form of (u+3)2(u+3)^{2}(u+3)2.(u+3)2=u2+6u+9(u+3)^{2} = u^{2} +6u +9(u+3)2=u2+6u+9

STEP 5

Now, substitute the expanded form of (u+3)2(u+3)^{2}(u+3)2 back into the original equation.2u2+14u+24=u2+u+92 u^{2}+14 u+24 = u^{2} +u +92u2+14u+24=u2+u+9

STEP 6

To solve for uuu, we need to bring all terms to one side of the equation. Subtract u2u^{2}u2, 6u6u6u, and 999 from both sides.2u2−u2+14u−6u+24−9=02 u^{2} - u^{2} +14 u -6u +24 -9 =02u2−u2+14u−6u+24−9=0

STEP 7

implify the equation.u2+u+15=0u^{2} +u +15 =0u2+u+15=0

STEP 8

Now, we need to factor the quadratic equation. We are looking for two numbers that multiply to 151515 and add to 888.(u+5)(u+3)=0(u +5)(u +3) =0(u+5)(u+3)=0

STEP 9

Set each factor equal to zero and solve for uuu.u+5=u +5 =u+5=u+3=u +3 =u+3=

STEP 10

olve for uuu in each equation.u=−5u = -5u=−5u=−3u = -3u=−3The solutions for uuu are −5-5−5 and −3-3−3.

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Solve for $u$ where $2 u^{2}+14 u+24=(u+3)^{2}$ . If multiple solutions, list them separated by commas.

Assumptions1. The equation is quadratic in nature. We are looking for real solutions for $u$

We are given the equation $2 u^{2}+14 u+24=(u+)^{2}$

First, we need to expand the right side of the equation $(u+3)^{2}$ .
$(u+3)^{2} = u^{2} +23u +3^{2}$

Calculate the expanded form of $(u+3)^{2}$ .
$(u+3)^{2} = u^{2} +6u +9$

Now, substitute the expanded form of $(u+3)^{2}$ back into the original equation.
$2 u^{2}+14 u+24 = u^{2} +u +9$

To solve for $u$ , we need to bring all terms to one side of the equation. Subtract $u^{2}$ , $6u$ , and $9$ from both sides.
$2 u^{2} - u^{2} +14 u -6u +24 -9 =0$

implify the equation.
$u^{2} +u +15 =0$

Now, we need to factor the quadratic equation. We are looking for two numbers that multiply to $15$ and add to $8$ .
$(u +5)(u +3) =0$

Set each factor equal to zero and solve for $u$ .
$u +5 =$ $u +3 =$

olve for $u$ in each equation.
$u = -5$ $u = -3$ The solutions for $u$ are $-5$ and $-3$ .