QuestionSolve the equation $x^{2}-8x-15=0$ and separate your answers with a comma.

Studdy Solution

STEP 1

Assumptions1. The equation is quadratic in nature, i.e., it is of the form $ax^ + bx + c =0$ . . The coefficients of the equation are real numbers.
3. The equation can be solved using the factorization method.

STEP 2

The given equation is $x^{2}-8x-15=0$ . We need to solve this equation for $x$ .

STEP 3

To solve the quadratic equation by factorization, we need to express the middle term as the sum of two terms whose coefficients multiply to give the product of the coefficient of $x^2$ (which is1 here) and the constant term (which is -15 here).

STEP 4

We need to find two numbers that add up to -8 (the coefficient of $x$ ) and multiply to -15 (the constant term). The numbers that satisfy these conditions are - and3.

STEP 5

Rewrite the equation by expressing -8x as the sum of -5x and3x.
$x^{2}-5x+3x-15=0$

STEP 6

Now, group the terms to factor by grouping.
$(x^{2}-5x)+(3x-15)=0$

STEP 7

Factor out the greatest common factor from each group.
$x(x-5)+3(x-5)=0$

STEP 8

Notice that $(x-5)$ is a common factor. Factor this out.
$(x-5)(x+3)=0$

STEP 9

Set each factor equal to zero and solve for $x$ .
$x-5= \quad or \quad x+3=$

STEP 10

olving the above equations gives the solutions to the original quadratic equation.
$x=5 \quad or \quad x=-3$ So, the solutions to the equation $x^{2}-8x-15=0$ are $x=5$ and $x=-3$ .

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QuestionSolve the equation $x^{2}-8x-15=0$ and separate your answers with a comma.

Studdy Solution

STEP 1

Assumptions1. The equation is quadratic in nature, i.e., it is of the form $ax^ + bx + c =0$ . . The coefficients of the equation are real numbers.
3. The equation can be solved using the factorization method.

STEP 2

The given equation is $x^{2}-8x-15=0$ . We need to solve this equation for $x$ .

STEP 3

To solve the quadratic equation by factorization, we need to express the middle term as the sum of two terms whose coefficients multiply to give the product of the coefficient of $x^2$ (which is1 here) and the constant term (which is -15 here).

STEP 4

We need to find two numbers that add up to -8 (the coefficient of $x$ ) and multiply to -15 (the constant term). The numbers that satisfy these conditions are - and3.

STEP 5

Rewrite the equation by expressing -8x as the sum of -5x and3x.
$x^{2}-5x+3x-15=0$

STEP 6

Now, group the terms to factor by grouping.
$(x^{2}-5x)+(3x-15)=0$

STEP 7

Factor out the greatest common factor from each group.
$x(x-5)+3(x-5)=0$

STEP 8

Notice that $(x-5)$ is a common factor. Factor this out.
$(x-5)(x+3)=0$

STEP 9

Set each factor equal to zero and solve for $x$ .
$x-5= \quad or \quad x+3=$

STEP 10

olving the above equations gives the solutions to the original quadratic equation.
$x=5 \quad or \quad x=-3$ So, the solutions to the equation $x^{2}-8x-15=0$ are $x=5$ and $x=-3$ .

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QuestionSolve the equation x2−8x−15=0x^{2}-8x-15=0x2−8x−15=0 and separate your answers with a comma.

Studdy Solution

STEP 1

Assumptions1. The equation is quadratic in nature, i.e., it is of the form ax+bx+c=0ax^ + bx + c =0ax+bx+c=0. . The coefficients of the equation are real numbers.3. The equation can be solved using the factorization method.

STEP 2

The given equation is x2−8x−15=0x^{2}-8x-15=0x2−8x−15=0. We need to solve this equation for xxx.

STEP 3

To solve the quadratic equation by factorization, we need to express the middle term as the sum of two terms whose coefficients multiply to give the product of the coefficient of x2x^2x2 (which is1 here) and the constant term (which is -15 here).

STEP 4

We need to find two numbers that add up to -8 (the coefficient of xxx) and multiply to -15 (the constant term). The numbers that satisfy these conditions are - and3.

STEP 5

Rewrite the equation by expressing -8x as the sum of -5x and3x.x2−5x+3x−15=0x^{2}-5x+3x-15=0x2−5x+3x−15=0

STEP 6

Now, group the terms to factor by grouping.(x2−5x)+(3x−15)=0(x^{2}-5x)+(3x-15)=0(x2−5x)+(3x−15)=0

STEP 7

Factor out the greatest common factor from each group.x(x−5)+3(x−5)=0x(x-5)+3(x-5)=0x(x−5)+3(x−5)=0

STEP 8

Notice that (x−5)(x-5)(x−5) is a common factor. Factor this out.(x−5)(x+3)=0(x-5)(x+3)=0(x−5)(x+3)=0

STEP 9

Set each factor equal to zero and solve for xxx.x−5=orx+3=x-5= \quad or \quad x+3=x−5=orx+3=

STEP 10

olving the above equations gives the solutions to the original quadratic equation.x=5orx=−3x=5 \quad or \quad x=-3x=5orx=−3So, the solutions to the equation x2−8x−15=0x^{2}-8x-15=0x2−8x−15=0 are x=5x=5x=5 and x=−3x=-3x=−3.

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QuestionSolve the equation $x^{2}-8x-15=0$ and separate your answers with a comma.

Assumptions1. The equation is quadratic in nature, i.e., it is of the form $ax^ + bx + c =0$ . . The coefficients of the equation are real numbers.
3. The equation can be solved using the factorization method.

The given equation is $x^{2}-8x-15=0$ . We need to solve this equation for $x$ .

To solve the quadratic equation by factorization, we need to express the middle term as the sum of two terms whose coefficients multiply to give the product of the coefficient of $x^2$ (which is1 here) and the constant term (which is -15 here).

We need to find two numbers that add up to -8 (the coefficient of $x$ ) and multiply to -15 (the constant term). The numbers that satisfy these conditions are - and3.

Rewrite the equation by expressing -8x as the sum of -5x and3x.
$x^{2}-5x+3x-15=0$

Now, group the terms to factor by grouping.
$(x^{2}-5x)+(3x-15)=0$

Factor out the greatest common factor from each group.
$x(x-5)+3(x-5)=0$

Notice that $(x-5)$ is a common factor. Factor this out.
$(x-5)(x+3)=0$

Set each factor equal to zero and solve for $x$ .
$x-5= \quad or \quad x+3=$

olving the above equations gives the solutions to the original quadratic equation.
$x=5 \quad or \quad x=-3$ So, the solutions to the equation $x^{2}-8x-15=0$ are $x=5$ and $x=-3$ .