For what values of x is x2-36 5x true

Category: What

Author: Hattie Newton

Published: 2021-03-06

Views: 994

What is the positive solution of x2 36 5x?

There is no real "solution" to this equation, as it is not an equation in the traditional sense. However, we can find the value of x that makes the equation true.

To do this, we first rewrite the equation in standard form:

x2 - 36 = 5x

We can then use the quadratic equation to solve for x:

x = (36 +/- sqrt(36^2 - 4*5*-36)) / (2*5)

x = (36 +/- sqrt(1296 + 1440)) / 10

x = (36 +/- sqrt(2736)) / 10

x = (36 +/- sqrt(1368)) / 10

x = (36 +/- 16.48) / 10

x = (52.48 or 19.52) / 10

x = 5.248 or 1.952

Thus, the value of x that makes the equation true is either 5.248 or 1.952.

What is the positive solution of x2 + 36 = 5x?

Assuming you are asking for the positive solution to the equation: x2 + 36 = 5x, we can solve for x by using algebra. We start by subtracting 36 from each side of the equation, which gives us: x2 - 36 = 5x - 36. Then, we subtract 5x from each side of the equation, which gives us: x2 - 5x - 36 = 0. To solve for x, we can use the quadratic equation. We set each side of the equation equal to 0 and then solve for x. The quadratic equation is: ax2 + bx + c = 0. In our equation, a = 1, b = -5, and c = -36. We plug these values into the quadratic equation and solve for x. We get: x = 6 or x = -2. Since we are looking for the positive solution, we can ignore the negative solution. Therefore, the positive solution to the equation is x = 6.

What is the positive value of x that satisfies the equation x2 + 36 = 5x?

There is no real answer to this question. It is impossible to solve for x without using imaginary numbers. However, we can approach this question by considering what would happen if we plug in different values for x. If we plug in x = 1, we get the equation 1 + 36 = 5, which is clearly not true. If we plug in x = 2, we get the equation 4 + 36 = 10, which is also not true. However, if we plug in x = 3, we get the equation 9 + 36 = 15, which is true. Therefore, we can conclude that the positive value of x that satisfies the equation x2 + 36 = 5x is 3. Even though we were unable to solve for x without using imaginary numbers, we were still able to find the positive value of x that satisfies the equation. This is a valuable lesson in itself, as it shows that sometimes there is more than one way to solve a problem. There is not always one "right" answer, and that is okay.

What is the value of x that makes the equation x2 + 36 = 5x true?

There is no real value of x that will make the equation x2 + 36 = 5x true. However, if we take a closer look at the equation, we can see that it is true for any value of x that is less than or equal to -6. This is because, when x is less than or equal to -6, the left side of the equation (x2 + 36) is always less than the right side of the equation (5x). So, while there is no one specific value of x that will make the equation true, we can say that it is true for any value of x that is less than or equal to -6.

What x value will produce a positive result when solving the equation x2 + 36 = 5x?

There is no real answer to this question because it is impossible to solve. The equation does not have a real solution because there is no value of x that will produce a positive result when solving the equation. The only way to solve this equation is to use imaginary numbers, which are not real.

What is the positive solution to the equation x2 + 36 = 5x?

There is no real solution to this equation, since there is no value of x that satisfies it. However, we can find an approximate solution by using the Quadratic Formula. If we plug in the values for a, b, and c from the equation, we get: x = -6 +/- sqrt(48) This gives us two possible values for x: -6+sqrt(48) and -6-sqrt(48). However, neither of these values is a real solution to the equation, since they both results in complex numbers. However, we can still use these values to approximately solve the equation. If we plug -6+sqrt(48) back into the equation, we get: (-6+sqrt(48))^2 + 36 = 5(-6+sqrt(48)) This simplifies to: 36+48-12+36 = -30+5sqrt(48) Which simplifies to: 84 = 5sqrt(48) And finally: 84/5 = sqrt(48) Therefore, the approximate solution to the equation is sqrt(48), or 6.6.

What is the value of x that will result in a positive answer when solving x2 + 36 = 5x?

There is no real value of x that will result in a positive answer when solving x^2 + 36 = 5x. This is because when you square x and add 36, you will always get a positive number, no matter what value x is. However, you can approximate the value of x that would result in a positive answer by using estimation. If you guess that x is around 3, then you can say that x^2 is 9 and 36 is 36. This means that 9 + 36 = 45, which is close to 5x (45 is closer to 50 than it is to 40). Therefore, 3 is a reasonable estimate for the value of x that would result in a positive answer when solving x^2 + 36 = 5x.

What is the positive value of x that will make the equation x2 + 36 = 5x true?

There is no real solution to this equation, as there is no real value of x that will make the equation true. However, we can use this equation to find the positive value of x that will make the equation true for complex numbers. If we set x = 3 + 2i, then we find that x2 + 36 = 5x becomes (3 + 2i)2 + 36 = 5(3 + 2i), which is true. Thus, the positive value of x that will make the equation x2 + 36 = 5x true is 3 + 2i.

What x value will result in a positive solution when solving the equation x2 + 36 = 5x?

When solving the equation, x2 + 36 = 5x, the x value that will result in a positive solution is x = 6. This is because when x = 6, the equation becomes 6^2 + 36 = 5(6), which simplifies to 36 + 36 = 30, and since 30 is a positive number, x = 6 is a positive solution.

What is the value of x that will produce a positive answer when solving the equation x2 +

There is no definitive answer to this question as it depends on the equation being solved. However, generally speaking, the value of x that will produce a positive answer when solving an equation is typically a positive number. This is because most equations are written in a form where the variable is on the left side of the equal sign and the constant term is on the right side. Therefore, in order for the equation to be balanced and produce a positive result, the value of x must be positive.

Related Questions

Factor the equation by grouping: x2 −5x−36 = 0 The solutions to this equation are x = −4 and x = 9.

To solve the equation algebraically: Now we can factor the expression by grouping. First, the expression needs to be rewritten as x^ {2}+ax+bx-36. x^2+x−5=0 Factorising this yields: x^2+1 = 0 x+(a−1) = 0 Solving for x yields: x = 1

Factor out x in the first and 4 in the second group. Factor out x in the first and 4 in the second group. Factor out common term x-9 by using distributive property. Factor out common term x − 9 by using distributive property. (x2−9x)+(4x−3 6)

The positive solution of x2 - 36 = 5x is 9.

By simplifying the right side of the equation, we can eliminate the exponents. In order to do this, we need to use the Square Root symbol (). The complete solution for 36 36 is 6 2 = 16.

X2-36 = 0 The easiest way to solve this equation is with algebra. After simplifying the equation, you can isolate the two terms and solve for x: x = 6 As you can see, x = 6 when X2-36 = 0.

To solve for X in a quadratic equation, use the quadratic formula. First, simplify the numerator (the coefficient on the left-hand side of the equation). This can be done by multiplying all of the terms on the left-hand side by negative one, canceling out common factors, and division by two: − 5 − 5 = 0 − 4 − 4 = −1 − 36 − 36 = 0 Next, simplify the denominator (the coefficient on the right-hand side of the equation). This can be done by dividing all of the terms on the right-hand side by negative one, canceling out common factors, and division by two:

(-5x)+36=0 (-5x)+9=0 x+36=0 x+9=36 x=27

To solve a quadratic equation with an order of 2, simply use the quadratic formula. The following equation can be solved using the quadratic formula: x = dfrac{-bpm sqrt{b^2 - 4ac}}{2a}.

The value of (2x2-x-36) is 0.

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