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allochka39001 [22]

3 years ago

5

The length of a rectangle is x feet more than its width. The area of the rectangle is 5x + 25. What is the width of the rectangl

e?

1 answer:

AVprozaik [17]3 years ago

6 0

Answer:

w = $l + \frac{25 - l^{2}}{l - 5}$

(Anyone can correct me if I'm wrong)

Step-by-step explanation:

Before we start the solving, we can make the following statements:

Area = l x w

Area = 5x + 25

therefore,

l x w = 5x + 25

Since the question states that the length is x more than the width, so we can make the following statement:

w = l + x

With this, we can substitute it to the first statement we made, l x w = 5x + 25,

l x (l + x) = 5x + 25

$l^{2}$ + lx = 5x + 25

lx - 5x = 25 - $l^{2}$

x(l - 5) = 25 - $l^{2}$

x = $\frac{25 - l^{2}}{l - 5}$

From this, we can find w by substituting it in the statement we made earlier, w = l + x,

w = $l + \frac{25 - l^{2}}{l - 5}$

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The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

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Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

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3 years ago