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3 years ago

Quais são os zeros(ou raízes) da função quadrática f(x) = x² + 3x - 4? Assinale a alternativa correta. *

Mathematics

1 answer:

faust18 [17]3 years ago

3 0

Given:

The quadratic function is:

$f(x)=x^2+3x-4$

To find:

The zeros of the given quadratic function.

Solution:

We have,

$f(x)=x^2+3x-4$

Splitting the middle term, we get

$f(x)=x^2+4x-x-4$

$f(x)=x(x+4)-1(x+4)$

$f(x)=(x+4)(x-1)$

For zeros, $f(x)=0$ .

$(x+4)(x-1)=0$

$x=-4,1$

Therefore, the correct option is (b).

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Sometimes life is tricky.

skad [1K]

thanks for the message !!

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3 years ago

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I will mark you the brainiest for the correct answer, please be correct, Have a good day and take care, thanks. ( VIEW THE IMAGE

makkiz [27]

Mistake in Line 2 : Was in the determining the values for a ,b,c

Mistake in Line 4 : Not taking the square roots of both sides.

Step-by-step explanation:

${a}^{2} + {b}^{2} = {c}^{2}$

First Mistake : Line 2

$a = x \\ b = 6 \\ c = 10$

Second Mistake : Line 4

$136 = {x}^{2} \\ not \\ 136 = x$

Correct Solution :

${x}^{2} + {6}^{2} = {10}^{2} \\ {x}^{2} = {10}^{2} - {6}^{2} \\ {x}^{2} = 100 - 36 \\$

${x}^{2} = 64 \\ \sqrt{ {x}^{2} } = \sqrt{64} \\ x = 6$

3 0

3 years ago

Consider the transpose of Your matrix A, that is, the matrix whose first column is the first row of A, the second column is the

Zarrin [17]

Answer:The system could have no solution or n number of solution where n is the number of unknown in the n linear equations.

Step-by-step explanation:

To determine if solution exist or not, you test the equation for consistency.

A system is said to be consistent if the rank of a matrix (say B ) is equal to the rank of the matrix formed by adding the constant terms(in this case the zeros) as a third column to the matrix B.

Consider the following scenarios:

(1) For example:Given the matrix A= $\left[\begin{array}{ccc}1&2\\3&4\end{array}\right]$ , to transpose A, exchange rows with columns i.e take first column as first row and second column as second row as follows:

Let A transpose be B.

∵B= $\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]$

the system Bx=0 can be represented in matrix form as:

$\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]$ $\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]$ = $\left[\begin{array}{ccc}0\\0\end{array}\right]$ ................................eq(1)

Now, to determine the rank of B, we work the determinant of the maximum sub-square matrix of B. In this case, B is a 2 x 2 matrix, therefore, the maximum sub-square matrix of B is itself B. Hence,

|B|=(1*4)-(3*2)= 4-6 = -2 i.e, B is a non-singular matrix with rank of order (-2).

Again, adding the constant terms of equation 1(in this case zeros) as a third column to B, we have $B_{0}$ :

$B_{0}$ = $\left[\begin{array}{ccc}1&3&0\\4&2&0\end{array}\right]$ . The rank of $B_{0}$ can be found by using the second column and third column pair as follows:

| $B_{0}$ |=(3*0)-(0*2)=0 i.e, $B_{0}$ is a singular matrix with rank of order 1.

Note: a matrix is singular if its determinant is = 0 and non-singular if it is $\neq$ 0.

Comparing the rank of both B and $B_{0}$ , it is obvious that

Rank of B $\neq$ Rank of $B_{0}$ since (-2)<1.

Therefore, we can conclude that equation(1) is inconsistent and thus has no solution.

(2) If B= $\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]$ is the transpose of matrix A= $\left[\begin{array}{ccc}-4&-8\\5&10\end{array}\right]$ , then

Then the equation Bx=0 is represented as:

$\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]$ $\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]$ = $\left[\begin{array}{ccc}0\\0\end{array}\right]$ ..................................eq(2)

|B|= (-4*10)-(5*(-8))= -40+40 = 0 i.e B has a rank of order 1.

$B_{0}$ = $\left[\begin{array}{ccc}-4&5&0\\-8&10&0\end{array}\right]$ ,

| $B_{0}$ |=(5*0)-(0*10)=0-0=0 i.e $B_{0}$ has a rank of order 1.

we can therefor conclude that since

rank B=rank $B_{0}$ =1, equation(2) is consistent and has 2 solutions for the 2 unknown ( $X_{1}$ and $X_{2}$ ).

Summary:

Given an equation Bx=0, transform the set of linear equations into matrix form as shown in equations(1 and 2).
Determine the rank of both the coefficients matrix B and $B_{0}$ which is formed by adding a column with the constant elements of the equation to the coefficient matrix.
If the rank of both matrix is same, then the equation is consistent and there exists n number of solutions(n is based on the number of unknown) but if they are not equal, then the equation is not consistent and there is no number of solution.

5 0

3 years ago

Which of the following is 6x²/27x simplified?

qaws [65]

Hi, The Answer is:

2x/9

Steps:

Simpilfy the expression:

6x/27

Then reduce the fraction with 3:

2x/9

That's your answer have a good day.

3 0

4 years ago

Read 2 more answers

Expand and simplify these 6 questions

zhuklara [117]

First one: divide multiply 2(1x) which would equal 2x then do 2*3 then u would subtract 5 which should get you to the simplified form 2x+1

Second one: do 3(1x) which would equal 3x then do 7*3 which would equal 21 than do plus 3x which should get you to the simplified form of 6x+21

Third one: Do 4(1x) which equals 4x than do 4*2 which equals 8 than plus eight which should get you to the simplest form of 4x+16

Fourth one: do 4(1x) which would equal 4x then do 4*1 which equals 4 than subtract 6 which should get you to the simplest form of 4x-2

Fifth one: do 2(3x) which equals 6x then do 2*2 which equals 4 than subtract 5x which should get you to the simplest form of x+4

Sixth one: do 5(1x) which equals 5x than do 5*-4 which equals -20 than add 10 which gets you to the simplest form of 5x-10

3 0

3 years ago