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3 years ago

Use the Quadratic formula to find all real zeros of the 2nd degree polynomial

Mathematics

1 answer:

schepotkina [342]3 years ago

4 0

Answer:

x ∈ {-5, -1}

Step-by-step explanation:

Here's the solution using the quadratic formula:

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-24\pm\sqrt{24^2-4\cdot 4\cdot 20}}{2\cdot 4}\\\\=\dfrac{-24\pm\sqrt{576-320}}{8}=\dfrac{-24\pm\sqrt{256}}{8}\\\\=\dfrac{-24\pm 16}{8}=-3\pm 2=\{-5,-1\}$

The real zeros are -5 and -1.

_____

There are many ways to check your answer. One of them is to look at the given quadratic, which has no changes of sign in its coefficients. (They are all positive.) That means there can be no positive real roots, so already you know that x=0.5 won't work.

Also, the constant in the quadratic is the product of the roots, For your roots, their product is -7/4, so even multiplying by 4 (the leading coefficient in the given quadratic), you don't get anything like 20.

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3 years ago

What is 10^-10 in decimal form?

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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.

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3 years ago

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faltersainse [42]

Answer:

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Step-by-step explanation:

1/3 of 12 is 4 so her son ate 4 cookies, your welcome

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3 years ago

Read 2 more answers