QuestionFactor the equation $x^{2}-2 x-35=0$ to find the values of $x$ .

Studdy Solution

STEP 1

Assumptions1. The equation is a quadratic equation in the form $ax^ + bx + c =0$ , where $a =1$ , $b = -$ , and $c = -35$ .

STEP 2

To solve the quadratic equation, we first need to factorize it. The factored form of a quadratic equation is $(x - p)(x - q) =0$ , where $p$ and $q$ are the roots of the equation.

STEP 3

We need to find two numbers that multiply to $-35$ (the value of $c$ ) and add to $-2$ (the value of $b$ ).

STEP 4

The two numbers that satisfy these conditions are $-7$ and $$ because $-7 \times = -35$ and $-7 + = -2$.

STEP 5

Therefore, the factored form of the equation is $(x - (-7))(x -5) =0$ .

STEP 6

implify the factored form of the equation.
$(x +)(x -5) =0$

STEP 7

Now, we can solve for $x$ by setting each factor equal to zero and solving for $x$ .
$x +7 =0$ $x -5 =0$

STEP 8

olve the first equation for $x$ .
$x = -7$

STEP 9

olve the second equation for $x$ .
$x =5$ So, the solutions to the equation $x^{2}-2 x-35=$ are $x = -7$ and $x =5$ .

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QuestionFactor the equation $x^{2}-2 x-35=0$ to find the values of $x$ .

Studdy Solution

STEP 1

Assumptions1. The equation is a quadratic equation in the form $ax^ + bx + c =0$ , where $a =1$ , $b = -$ , and $c = -35$ .

STEP 2

To solve the quadratic equation, we first need to factorize it. The factored form of a quadratic equation is $(x - p)(x - q) =0$ , where $p$ and $q$ are the roots of the equation.

STEP 3

We need to find two numbers that multiply to $-35$ (the value of $c$ ) and add to $-2$ (the value of $b$ ).

STEP 4

The two numbers that satisfy these conditions are $-7$ and $$ because $-7 \times = -35$ and $-7 + = -2$.

STEP 5

Therefore, the factored form of the equation is $(x - (-7))(x -5) =0$ .

STEP 6

implify the factored form of the equation.
$(x +)(x -5) =0$

STEP 7

Now, we can solve for $x$ by setting each factor equal to zero and solving for $x$ .
$x +7 =0$ $x -5 =0$

STEP 8

olve the first equation for $x$ .
$x = -7$

STEP 9

olve the second equation for $x$ .
$x =5$ So, the solutions to the equation $x^{2}-2 x-35=$ are $x = -7$ and $x =5$ .

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QuestionFactor the equation x2−2x−35=0x^{2}-2 x-35=0x2−2x−35=0 to find the values of xxx.

Studdy Solution

STEP 1

Assumptions1. The equation is a quadratic equation in the form ax+bx+c=0ax^ + bx + c =0ax+bx+c=0, where a=1a =1a=1, b=−b = -b=−, and c=−35c = -35c=−35.

STEP 2

To solve the quadratic equation, we first need to factorize it. The factored form of a quadratic equation is (x−p)(x−q)=0(x - p)(x - q) =0(x−p)(x−q)=0, where ppp and qqq are the roots of the equation.

STEP 3

We need to find two numbers that multiply to −35-35−35 (the value of ccc) and add to −2-2−2 (the value of bbb).

STEP 4

The two numbers that satisfy these conditions are −7-7−7 and $$ because $-7 \times = -35$ and $-7 + = -2$.

STEP 5

Therefore, the factored form of the equation is (x−(−7))(x−5)=0(x - (-7))(x -5) =0(x−(−7))(x−5)=0.

STEP 6

implify the factored form of the equation.(x+)(x−5)=0(x +)(x -5) =0(x+)(x−5)=0

STEP 7

Now, we can solve for xxx by setting each factor equal to zero and solving for xxx.x+7=0x +7 =0x+7=0x−5=0x -5 =0x−5=0

STEP 8

olve the first equation for xxx.x=−7x = -7x=−7

STEP 9

olve the second equation for xxx.x=5x =5x=5So, the solutions to the equation x2−2x−35=x^{2}-2 x-35=x2−2x−35= are x=−7x = -7x=−7 and x=5x =5x=5.

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QuestionFactor the equation $x^{2}-2 x-35=0$ to find the values of $x$ .

Assumptions1. The equation is a quadratic equation in the form $ax^ + bx + c =0$ , where $a =1$ , $b = -$ , and $c = -35$ .

To solve the quadratic equation, we first need to factorize it. The factored form of a quadratic equation is $(x - p)(x - q) =0$ , where $p$ and $q$ are the roots of the equation.

We need to find two numbers that multiply to $-35$ (the value of $c$ ) and add to $-2$ (the value of $b$ ).

The two numbers that satisfy these conditions are $-7$ and $$ because $-7 \times = -35$ and $-7 + = -2$.

Therefore, the factored form of the equation is $(x - (-7))(x -5) =0$ .

implify the factored form of the equation.
$(x +)(x -5) =0$

Now, we can solve for $x$ by setting each factor equal to zero and solving for $x$ .
$x +7 =0$ $x -5 =0$

olve the first equation for $x$ .
$x = -7$

olve the second equation for $x$ .
$x =5$ So, the solutions to the equation $x^{2}-2 x-35=$ are $x = -7$ and $x =5$ .