Solved on Sep 27, 2023

Find values of $x$ such that $(x+7)(x-5)=13$ . The solution set is $x=2, x=-12$ .

STEP 1

Assumptions1. The given equations are $y_{1}=(x+7)$ , $y_{}=(x-5)$ , and $y_{1} y_{}=13$ . . We are asked to find all values of $x$ that satisfy these conditions.

STEP 2

From the given conditions, we know that $y_{1} y_{2}=13$ . We can substitute $y_{1}=(x+7)$ and $y_{2}=(x-5)$ into this equation.
$y_{1} y_{2} = (x+7)(x-5)$

STEP 3

Now, we set the equation equal to13.
$(x+7)(x-5) =13$

STEP 4

Next, we expand the left side of the equation.
$x^2 +2x -35 =13$

STEP 5

Subtract13 from both sides of the equation to set it equal to zero.
$x^2 +2x -48 =0$

STEP 6

Now, we factor the quadratic equation.
$(x -6)(x +8) =0$

STEP 7

Setting each factor equal to zero gives the solutions for $x$ .
$x -6 =0 \quad \text{or} \quad x + =0$

STEP 8

olving each equation gives the solutions for $x$ .
$x =6 \quad \text{or} \quad x = -8$ The solution set is $\{6, -8\}$ .

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Find values of $x$ such that $(x+7)(x-5)=13$ . The solution set is $x=2, x=-12$ .

STEP 1

Assumptions1. The given equations are $y_{1}=(x+7)$ , $y_{}=(x-5)$ , and $y_{1} y_{}=13$ . . We are asked to find all values of $x$ that satisfy these conditions.

STEP 2

From the given conditions, we know that $y_{1} y_{2}=13$ . We can substitute $y_{1}=(x+7)$ and $y_{2}=(x-5)$ into this equation.
$y_{1} y_{2} = (x+7)(x-5)$

STEP 3

Now, we set the equation equal to13.
$(x+7)(x-5) =13$

STEP 4

Next, we expand the left side of the equation.
$x^2 +2x -35 =13$

STEP 5

Subtract13 from both sides of the equation to set it equal to zero.
$x^2 +2x -48 =0$

STEP 6

Now, we factor the quadratic equation.
$(x -6)(x +8) =0$

STEP 7

Setting each factor equal to zero gives the solutions for $x$ .
$x -6 =0 \quad \text{or} \quad x + =0$

STEP 8

olving each equation gives the solutions for $x$ .
$x =6 \quad \text{or} \quad x = -8$ The solution set is $\{6, -8\}$ .

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Find values of xxx such that (x+7)(x−5)=13(x+7)(x-5)=13(x+7)(x−5)=13. The solution set is x=2,x=−12x=2, x=-12x=2,x=−12.

STEP 1

Assumptions1. The given equations are y1=(x+7)y_{1}=(x+7)y1​=(x+7), y=(x−5)y_{}=(x-5)y​=(x−5), and y1y=13y_{1} y_{}=13y1​y​=13. . We are asked to find all values of xxx that satisfy these conditions.

STEP 2

From the given conditions, we know that y1y2=13y_{1} y_{2}=13y1​y2​=13. We can substitute y1=(x+7)y_{1}=(x+7)y1​=(x+7) and y2=(x−5)y_{2}=(x-5)y2​=(x−5) into this equation.y1y2=(x+7)(x−5)y_{1} y_{2} = (x+7)(x-5)y1​y2​=(x+7)(x−5)

STEP 3

Now, we set the equation equal to13.(x+7)(x−5)=13(x+7)(x-5) =13(x+7)(x−5)=13

STEP 4

Next, we expand the left side of the equation.x2+2x−35=13x^2 +2x -35 =13x2+2x−35=13

STEP 5

Subtract13 from both sides of the equation to set it equal to zero.x2+2x−48=0x^2 +2x -48 =0x2+2x−48=0

STEP 6

Now, we factor the quadratic equation.(x−6)(x+8)=0(x -6)(x +8) =0(x−6)(x+8)=0

STEP 7

Setting each factor equal to zero gives the solutions for xxx.x−6=0orx+=0x -6 =0 \quad \text{or} \quad x + =0x−6=0orx+=0

STEP 8

olving each equation gives the solutions for xxx.x=6orx=−8x =6 \quad \text{or} \quad x = -8x=6orx=−8The solution set is {6,−8}\{6, -8\}{6,−8}.

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Find values of $x$ such that $(x+7)(x-5)=13$ . The solution set is $x=2, x=-12$ .

Assumptions1. The given equations are $y_{1}=(x+7)$ , $y_{}=(x-5)$ , and $y_{1} y_{}=13$ . . We are asked to find all values of $x$ that satisfy these conditions.

From the given conditions, we know that $y_{1} y_{2}=13$ . We can substitute $y_{1}=(x+7)$ and $y_{2}=(x-5)$ into this equation.
$y_{1} y_{2} = (x+7)(x-5)$

Now, we set the equation equal to13.
$(x+7)(x-5) =13$

Next, we expand the left side of the equation.
$x^2 +2x -35 =13$

Subtract13 from both sides of the equation to set it equal to zero.
$x^2 +2x -48 =0$

Now, we factor the quadratic equation.
$(x -6)(x +8) =0$

Setting each factor equal to zero gives the solutions for $x$ .
$x -6 =0 \quad \text{or} \quad x + =0$

olving each equation gives the solutions for $x$ .
$x =6 \quad \text{or} \quad x = -8$ The solution set is $\{6, -8\}$ .