# A student performed the following steps to find the solution to the equationx2 - 2x - 15 = 0. Where did the student go

###### Question:

Step 1. Factor the polynomial into (x+3) and (x - 5)

Step 2. X+ 3 = 0 and XI 5 = 0

Step 3. x = 3 and x = -5

A. in Step 3

B. The student did not make any mistakes, the solution is correct

c. in Step 1

D. in Step 2

## Answers

A quadratic equation is one of the form ax2 + bx + c = 0, where a, b, and c are numbers, and a is not equal to 0. So I think x = 4 but not 100%

Sorry if I'm wrong but I hope this helps

the solutions are {1, -3/2}

Step-by-step explanation:

The given equation 2x^2 + x - 1 = 2 simplifies to 2x^2 + x - 3 = 0 if we subtract 2 from both sides.

We are told that -3/2 is a solution and must find the other solution To do this, use synthetic division; if the remainder is zero, that tells us that -3/2 is actually a solution, and we can obtain the second solution easily, as follows:

-3/2 / 2 1 -3

-3 3

2 -2 0

Since the remainder is 0, we have verified that -3/2 is a solution. The coefficients of the other factor of the original equation are 2 and -2, and so this other factor is 2x - 2, or 2(x - 1). Setting this result equal to zero yields x = 1.

Thus the solutions are {1, -3/2}.

2x times 2

Step-by-step explanation:

[tex]2 {x}^{2} + x - 1 = 2[/tex]

Subtract sides -2

[tex]2 {x}^{2} + x - 1 - 2 = 2 - 2[/tex]

[tex]2 {x}^{2} + x - 3 = 0[/tex]

[tex](x - 1)(2x + 3) = 0[/tex]

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[tex]x - 1 = 0[/tex]

[tex]x = 1[/tex]

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[tex]2x + 3 = 0[/tex]

[tex]x = - \frac{3}{2} \\[/tex]

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Done♥️♥️♥️♥️♥️

Only the 2 is negative. The 2nd step is incorrect.

Step-by-step explanation:

In this equation, she did the 2nd step wrong because when you use the factors of 8, -4 and -2 just get you -6, not -2. So, you would have a -4 and a positive 2.

Option 4th is correct

x = 2

Step-by-step explanation:

Using logarithmic rule:

[tex]\log_b x = \log_b y[/tex]

then; x = y

Given the equation:

[tex]\log_3 (x^2+6x) = \log_3 (2x+12)[/tex]

Apply the rule:

[tex]x^2+6x = 2x+12[/tex]

Subtract 2x from both sides we have;

[tex]x^2+4x =12[/tex]

Subtract 12 from both sides we have;

[tex]x^2+4x-12=0[/tex]

⇒[tex]x^2+6x-2x-12 = 0[/tex]

⇒[tex]x(x+6)-2(x+6)= 0[/tex]

⇒[tex](x+6)(x-2)= 0[/tex]

by zero product property we have:

x+6 =0 and x-2 = 0

⇒x= -6 and x = 2

Since, at x = -6 it does not satisfy the given equation.

Therefore, the only solution to the given equation is, x = 2

Option 4th is correct

x = 2

Step-by-step explanation:

Using logarithmic rule:

[tex]\log_b x = \log_b y[/tex]

then; x = y

Given the equation:

[tex]\log_3 (x^2+6x) = \log_3 (2x+12)[/tex]

Apply the rule:

[tex]x^2+6x = 2x+12[/tex]

Subtract 2x from both sides we have;

[tex]x^2+4x =12[/tex]

Subtract 12 from both sides we have;

[tex]x^2+4x-12=0[/tex]

⇒[tex]x^2+6x-2x-12 = 0[/tex]

⇒[tex]x(x+6)-2(x+6)= 0[/tex]

⇒[tex](x+6)(x-2)= 0[/tex]

by zero product property we have:

x+6 =0 and x-2 = 0

⇒x= -6 and x = 2

Since, at x = -6 it does not satisfy the given equation.

Therefore, the only solution to the given equation is, x = 2

positive 3/2

Step-by-step explanation:

see explanation

Step-by-step explanation:

Given

2x² + x - 1 = 2 ( subtract 2 from both sides )

2x² + x - 3 = 0

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term

product = 2 × - 3 = - 6 and sum = + 1

The factors are - 2 and + 3

Use these factors to split the x- term

2x² - 2x + 3x - 3 = 0 ( factor the first/second and third/fourth terms )

2x(x - 1) + 3(x - 1) = 0 ← factor out (x - 1) from each term

(x - 1)(2x + 3) = 0

Equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

2x + 3 = 0 ⇒ 2x = - 3 ⇒ x = - [tex]\frac{3}{2}[/tex]

the answer is option 4: x=2