All categories

3 years ago

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

Mathematics

1 answer:

Wewaii [24]3 years ago

4 0

Answer:

Required positive solution of the given quadratic equation is 9.

Step-by-step explanation:

Given Equation,

x² - 36 = 5x

We need to find positive solution of the given equation.

We solve the given quadratic equation using middle term split method.

x² - 36 = 5x

x² - 5x - 36 = 0

x² - 9x + 4x - 36 = 0

x( x - 9 ) + 4( x - 9 ) = 0

( x - 9 )( x + 4 ) = 0

x - 9 = 0 and x + 4 = 0

x = 9 and x = -4

Therefore, Required positive solution of the given quadratic equation is 9.

You might be interested in

The sale price of a new compact disc player is $212 at a local discount store. At the store where this sale is going on, each ne

love history [14]

Answer:

6 CDs

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Andrea ran 4 a miles on

Nikitich [7]

Answer:9

Step-by-step explanation:5+4=9

3 0

3 years ago

Read 2 more answers

Write a system of linear equations for the graph below

Oliga [24]

1q+2x-1 liner #1 liner # is 18% one

5 0

4 years ago

What is the constant of 8x^2-7x+6

ruslelena [56]

Answer:

6 is the constant

Step-by-step explanation:

8x^2-7x+6

The constant is the term that does not change ( the one with out the variable)

6 is the constant

5 0

3 years ago

A tank holds 300 gallons of water and 100 pounds of salt. A saline solution with concentration 1 lb salt/gal is added at a rate

Shkiper50 [21]

The amount of salt in the tank changes with rate according to

$Q'(t)=\left(1\dfrac{\rm lb}{\rm gal}\right)\left(4\dfrac{\rm gal}{\rm min}\right)-\left(\dfrac{Q(t)}{300+(4-1)t}\dfrac{\rm lb}{\rm gal}\right)\left(1\dfrac{\rm gal}{\rm min}\right)$

$\implies Q'+\dfrac Q{300+3t}=4$

which is a linear ODE in $Q(t)$ . Multiplying both sides by $(300+3t)^{1/3}$ gives

$(300+3t)^{1/3}Q'+(300+3t)^{-2/3}Q=4(300+3t)^{1/3}$

so that the left side condenses into the derivative of a product,

$\big((300+3t)^{1/3}Q\big)'=4(300+3t)^{1/3}$

Integrate both sides and solve for $Q(t)$ to get

$(300+3t)^{1/3}Q=(300+3t)^{4/3}+C$

$\implies Q(t)=300+3t+C(300+3t)^{-1/3}$

Given that $Q(0)=100$ , we find

$100=300+C\cdot300^{-1/3}\implies C=-200\cdot300^{1/3}$

and we get the particular solution

$Q(t)=300+3t-200\cdot300^{1/3}(300+3t)^{-1/3}$

$\boxed{Q(t)=300+3t-2\cdot100^{4/3}(100+t)^{-1/3}}$

5 0

3 years ago