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3 years ago

an integer is one more than four times another. if the product of two integers is 39, then find the integers

Mathematics

2 answers:

Sholpan [36]3 years ago

8 0

Sorry if its illegible, you can ask if its not clear

strojnjashka [21]3 years ago

3 0

Answer:

The integers are 3 and 13

Step-by-step explanation:

Let the integers are x and y.

Then from the given directions we can frame two equations.

$x=4y+1.........(1)\\\\xy=39..........(2)\\\\$

From equation 2, the value of x is $x=\frac{39}{y}$

On substituting this value in equation 1, we get

$\frac{39}{y}=4y+1\\4y^2+y-39=0$

Applying the quadratic formula, we get

$y_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:4\left(-39\right)}}{2\cdot \:4}\\\\y=3,\:y=-\frac{13}{4}$

Among these two values, 3 is an integer.

Hence, y= 3

The value of x is x= 39/3 = 13

Therefore, the integers are 3 and 13

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3 years ago

Read 2 more answers

Evaluate the function for the given values to determine if the value is a root. p(−2) = p(2) = The value is a root of p(x).

bija089 [108]

Note: Since you missed to mention the the expression of the function $p(x)$ . After a little research, I was able to find the complete question. So, I am assuming the expression as $p(x)=x^4-9x^2-4x+12$ and will solve the question based on this assumption expression of $p(x)$ , which anyways would solve your query.

Answer:

$p\left(-2\right)=0$

Therefore, $x=-2$ is a root of the polynomial $p(x)=x^4-9x^2-4x+12$

$p\left(2\right)=-16$

Therefore, $x=2$ is not a root of the polynomial $p(x)=x^4-9x^2-4x+12$

Step-by-step explanation:

As we know that for any polynomial let say $p(x)$ , $c$ is the root of the polynomial if $p(c)=0$ .

In order to find which of the given values will be a root of the polynomial, $p(x)=x^4-9x^2-4x+12$ , we must have to evaluate $p(x)$ for each of these values to determine if the output of the function gets zero.

So,

Solving for $p\left(-2\right)$