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True [87]
3 years ago
6

Find the solution(s) to x^2- 14x + 49 = 0.

Mathematics
1 answer:
malfutka [58]3 years ago
3 0

Answer:

\boxed{C. \: x = 7 \: only}

Step-by-step explanation:

=  >  {x}^{2}  - 14x + 49 = 0 \\  \\  =  >  {x}^{2} - (7 + 7)x + 49 = 0 \\  \\  =  >  {x}^{2}  - 7x - 7x + 49 = 0 \\  \\  =  > x(x - 7) - 7(x - 7) = 0 \\  \\  =  > (x - 7)(x - 7) = 0 \\  \\  =  >  {(x - 7)}^{2}  = 0 \\  \\  =  > x - 7 = 0 \\  \\  =  > x = 7

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The area of a rectangle is 45 cm2. Two squares are constructed such that two adjacent sides of the rectangle are each also the s
Zielflug [23.3K]

Answer:

The lengths of sides of squares are 5 cm and 9 cm.

Step-by-step explanation:

Let the sides of rectangle be x and y .

Area of rectangle = Length × Breadth

Given : Area of rectangle = 45 cm².

⇒ x × y  = 45  

⇒ x=\frac{45}{y}  ........(1)

Two squares are constructed such that two adjacent sides of the rectangle

so squares have side x and y .

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Area of square with side x = x²

Area of square with side y = y²

Also, The combined area of the two squares is 106 cm².

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From (1) put value of x , we get,

(\frac{45}{y})^2+y^2=106

Solving for y ,

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2025+y^4=106y^2  

y^4-106y^2+2025=0  

y^4-81y^2-25y^2+2025=0  

y^2(y^2-81)-25(y^2-81)=0  

(y^2-25)(y^2-81)=0

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on solving we get y = 5 and y = 9

Also, x can be find by putting in (1),

⇒ x=\frac{45}{5}=9 and ⇒ x=\frac{45}{9}=5  

⇒ x = 9 and x = 5

Thus, lengths of sides of squares are 5 cm and 9 cm.

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Answer:

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