Question

A. Solve by Factoring and Completing the Square

Juan Dela Cruz has a rectangular center table. He wants to cover it with a table cloth having a length of 2m longer than its width. The area is 24m². What will be the length and width of the table cloth?

B. Solve by Quadratic Formula and Extracting Square Roots

The utility boy of a certain school followed the request of the principal to divide the 4 m water pipe into two parts so that the sum of their reciprocal is 1. What will be the measures of the two parts of the water pipe?

pa sagot po please​

Answer 1

Sorry Friend,

I know only (B)

Let x be the first part of the pipe

4 - x be the second part.

(1/x) + (1/4-x)=1

(4-x)+(x)=x (4-x)

4=4x-x^2

x^2-4x+4=0

(x-2) (x-2)=0

x=0.

I hope it's will help you.

please mark me as Brainliest.

Answer 2

Given:

A) Length of table cloth = 2m longer than width

Area of table cloth = 24 m²

B) Length of water pipe = 4m

Water pipe divided into 2 parts

To find:

A) Length of cloth.

Width of cloth.

B) Measures of the two parts of pipe.

Solution:

A) Let $l$ be the length of the cloth and $b$ be the width of the table cloth.

Then according to the question, $l=2+b$

Since, this cloth is rectangular in shape, its area is given by:

$Area=length*breadth$

Let $A$ be the area of cloth which is given as $24m^{2}$.

$A=l*b$

$24=(2+b)b$

$24=2b+b^{2}$

To solve by factoring and completing the squares, first divide both sides with the coefficient of $b^{2}$ which is $1$.

Next take half of the coefficient of $b$ which is $\frac{2}{2}=1$.

Square this number, $1^{2} =1$

Add this number on both sides.

$24+1=2b+b^{2} +1$

$b^{2}+2b+1=24+1$

Factorize the LHS.

$(b+1)^{2}=25$

Taking square root on both sides.

$b+1=\pm5$

On solving,

$b+1=5$ or $b+1=-5$

$b=5-1$ or $b=-5-1$

$b=4$ or $b=-6$ (rejected, length can never be negative)

Hence, width of rectangular table cloth is $4m$.

Length = $2+b=2+4=6m$

Length of table cloth is $6m$.

Hence, length and width of the rectangular table cloth is $6m$ and $4m$ respectively.

B) The water pipe of length $4m$ is divided into 2 parts such that sum of their reciprocals is $1$.

This pipe can be divided into a part having length $1m$ and another part having length $3m.$

Sum of their reciprocals = $\frac{1}{1}+\frac{1}{3}=\frac{1+3}{3}=\frac{4}{3} \neq 1$

This pipe can also be divided into 2 parts each having length $2m$.

Sum of their reciprocals = $\frac{1}{2}+\frac{1}{2}=1$

Hence, the pipe is to be divided into two parts each having length $2m$.

A) Length and width of the rectangular table cloth is $6m$ and $4m$

respectively.

B) The pipe is to be divided into two parts each having length $2m$.