All categories

3 years ago

Solve the equation k^2-6k=11 by completing the square

Mathematics

2 answers:

Alex787 [66]3 years ago

3 0

Answer:

k = ±2√5 + 3

Step-by-step explanation:

k² - 6k + 9 = 11 + 9
(k - 3)² = 20
k - 3 = √20
k - 3 = ±2√5
k = ±2√5 + 3

VMariaS [17]3 years ago

3 0

<h2>Answer:</h2>

$k=3+2\sqrt5\ or\ k=3-2\sqrt5$

<h2>Step-by-step explanation:</h2>

$k^2-6k=11$

_______________________________________________

<h3>Add one term in order to complete the square</h3>

$k^2-6k+(6\times\frac{1}{2})^2=11+(6\times\frac{1}{2})^2$

________________________________________________

<h3>Calculate</h3>

$k^2-6k+3^2=11+3^2$

_________________________________________________

<h3>Factor the expression using $a^2\pm 2ab+b^2=(a\pm b)^2$ </h3>

$(k-3)^2=11+3^2$

__________________________________________________

<h3>Calculate the power</h3>

$(k-3)^2=11+9$

___________________________________________________

<h3>Calculate the sum or difference</h3>

$(k-3)^2=20$

______________________________________________________

<h3>Split into two equations</h3>

$k-3=\sqrt{20}\ or\ k-3=-\sqrt{20}$

________________________________________________________

<h3>Move variables to the left side of the equation:</h3>

$(k-3=\sqrt{20})$

$k=\sqrt{20}+3\\ k=2\sqrt{5}+3$

____________________________________________________

<h3>Move variables to the left side of the equation:</h3>

$(k-3=-\sqrt{20})$

$k=\sqrt{20}+3\\ k=-2\sqrt{5}+3$

_____________________________________________________

$k=2\sqrt{5}+3\ or\ k=-2\sqrt{5}+3$

_____________________________________________________

<h3>Find the union of the solutions</h3>

$k=2\sqrt{5}+3\ or\ k=-2\sqrt{5}+3$

_____________________________________________________

I hope this helps you

:)

<h3 />

You might be interested in

Find the distance between (15, -7) and (-20, -5) The distance is __ units

Nimfa-mama [501]

Answer: The distance between the two points in exact form is

√1229 and the decimal form is 35.05709628...

4 0

3 years ago

Paige and Cindy have been saving money. Paige started with $350 in her

ehidna [41]

Answer:

Thus the prime factors of the given numbers can be only 2, 3, 5, and 7. ... most 3 + 1 + 1 + 1 = 6 numbers, we are left with at least three numbers among the nine ... points, and b by blue points (for these inequalities to hold we might have to ... 8. The constant function f (x) = k, where k is a positive integer, is the only possible.

Step-by-step explanation:

4 0

3 years ago

What is -7x+2y=-23 y=-8x+23

Zielflug [23.3K]

Steps:I will try to solve your system of equations.
y=−8x+23;−7x+2y=−23
Step: Solve y=−8x+23 for y: Step:
Substitute−8x+23 for y in −7x+2y=−23:
−7x+2y=−23−7x+2(−8x+23)=−23−23x+46=−23(Simplify both sides of the equation)−23x+46+−46=−23+−46(Add -46 to both sides)−23x=−69−23x−23=−69−23(Divide both sides by -23)x=3Step: Substitute 3 for x in y=−8x+23:y=−8x+23y=(−8)(3)+23y=−1(Simplify both sides of the equation)
Answer:y=−1 and x=3

6 0

3 years ago

There are some Nichols dimes and quarters in a large piggy bank for every two nickels there a sweetheart forever to Diane's ther

maxonik [38]

Question is not proper, Proper question is given below.

There are some nickels, dimes, and quarters in a large piggy bank. For every 2 nickels, there are 3 dimes. For every 2 dimes, there are 5 quarters. There are 500 coins in all. How many nickels, dimes, and quarters are in the piggy bank? How much are the coins in the piggy bank worth all together?

Answer:

There are 80 nickels, 120 dimes and 300 quarters in the piggy bank.

The piggy bank is worth $91.

Step-by-step explanation:

Let Number of Nickels be represented by 'n'.

Let Number of Dimes be represented by 'd'.

Let of Quarters be represented by 'q'.

Now Given:

For every 2 nickels, there are 3 dimes.

2 n = 3 d

1 n = number of dimes.

We will use Unitary method to find the same.

Number of 1 dimes d = $\frac{3}{2}n$

Also Given:

For every 2 dimes, there are 5 quarters.

2 d = 5 q

1 d = number of quarters

We will use Unitary method to find the same.

Number of 1 quarter q = $\frac{5}{2}d$

Also Given:

Total Number of coins = 500

$n+d+q=500$

Now d = $\frac{3}{2}n$

AND q = $\frac{5}{2}d = \frac{5}{2} \times\frac{3}{2}n = \frac{15}{4}n$

So Substituting the values we get;

$n+\frac{3}{2}n+\frac{15}{4}n=500$

Taking LCM we get making denominator as 4.

$\frac{4}{4}n+\frac{3\times2}{2\times 2}n+\frac{15\times 1}{4\times 1}n=500\\\\\frac{4}{4}n+\frac{6}{4}n+\frac{15}{4}n=500\\\\\frac{4n+6n+15n}{4}=500\\\\\frac{25n}4 =500\\\\25n = 500\times 4\\\\25n =2000\\\\n=\frac{2000}{25}=80$