Question

$\huge{\underline{\color{teal}{\textsf{\textbf{Question~~~~}}}}}$
Q. determine x = 5 is the root of given equation or not
$\sqrt{2x {}^{2} + 4x - 5} - \sqrt{x {}^{2} - 4x + 4 } = \sqrt{1 - 12x + 3x {}^{2} }$
Q. the difference between 2 number is 5 and the difference is between their squares is 55 find the largest number
• 9
• 10
• 11
• 12

Q. solve the following quadratic equation
$\frac{x - 1}{x - 2} - \frac{x - 2}{x - 3} = \frac{ x - 5}{x - 6} - \frac{x - 6}{x - 7}$
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Answer 1

Step-by-step explanation:

Answers :-

1)Given equation is

[√(2x^2+4x-5)]-[√(x^2-4x+4)]=√(1-12x+3x^2)

If 5 is a root then it satisfies the given equation i.e. LHS = RHS

Put x = 5 then

LHS = [√(2x^2+4x-5)]-[√(x^2-4x+4)]

=> [√(2(5)^2+4(5)-5)]-[√{(5)^2-4(5)+4}]

=> [√(2(25)+20-5)]-[√{25-20+4}]

=> √(50+20-5)-√(25-20+4)

=>√(70-5)-√(29-20)

=> √65 -√9

=>√65-3

RHS:-√(1-12x+3x^2)

=>√[(1-12(5)+3(5)^2]

=>√(1-60+75)

=>√(76-60)

=>√16

=>4

LHS≠RHS

So 5 is not a root of the given equation.

2)

Let the two numbers be X and Y

Let X> Y

Large number = X

Small number = Y

Their difference = 5

=>X-Y = 5-------(1)

Difference between their squares = 55

=>X^2-Y^2 = 55

=>(X+Y)(X-Y)=55

=>(X+Y)(5) = 55

=>X+Y = 55/5

=>X+Y = 11------(2)

adding (1)&(2)

X-Y = 5

X+Y = 11

(+)

________

2X +0=16

_________

=>2X = 16

X=16/2

=>X = 8

and from (2)

8+Y=11

=>Y = 11-8

Y=3

Large number = 8

Check:-

8-3 = 5

8^2-3^2

=>64-9

=55

Verified the given relations

3)

Given that:

(x-1)/(x-2)- (x-2)/(x-3) = (x-5)/(x-6) -(x-6)/(x-7)

[(x-1)(x-3)-(x-2)(x-2)]/(x-2)(x-3)

=[(x-5)(x-7)-(x-6)(x-6)]/(x-6)(x-7)

[x^2-x-3x+3-x^2+4x-4]/(x^2-2x-3x+6) =[x^2-5x-7x+35-x^2+12x-36)/(x^2-6x-7x+42)

[x^2-x^2-4x+4x+3-4]/(x^2-5x+6)= [x^2-x^2-12x+12x-1]/(x^2-13x+42)

-1/(x^2-5x+6) = -1/(x^2-13x+42)

=>1/(x^2-5x+6) = 1/(x^2-13x+42)

=> x^2-13x+42 = x^2-5x+6

=>-13x+42 = -5x+6

=>-13x+5x=6-42

=>-8x=-36

=>x = -36/-8

=>x = 9/2

The value of x = 9/2

Answer 2

Question:-

Determine whether x = 5 is the root of the given equation.

$\sf{\sqrt{2x^2 + 4x - 5} - \sqrt{x^2 - 4x + 4} = \sqrt{1 - 12x + 3x^2}}$

Solution:-

Putting x = 5 in the given quadratic equation.

$=\sf{\sqrt{2 \times (5)^2 + 4 \times 5 - 5} - \sqrt{(5)^2 - 4 \times 5 + 4} = \sqrt{1 - 12\times 5 + 3\times (5)^2}}$

$=\sf{\sqrt{2 \times 25 + 20 - 5} - \sqrt{25 - 20 + 4} = \sqrt{1 - 60 + 3\times 25}}$

$=\sf{\sqrt{50 + 20 - 5} - \sqrt{29 - 20} = \sqrt{1 - 60 + 75}}$

$= \sf{\sqrt{70 - 5} - \sqrt{9} = \sqrt{76 - 60}}$

$=\sf{\sqrt{65} - 3 = \sqrt{16}}$

$= \sf{\sqrt{65} - 3 = 4}$

Here,

$\sf{LHS \neq RHS}$

$\sf{\therefore \sqrt{2x^2 + 4x - 5} - \sqrt{x^2 - 4x + 4} = \sqrt{1 - 12x + 3x^2}\:is\:not\:the\:root\:of\:the\:equation}$

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Question:-

The difference between 2 number is 5 and the difference between their squares is 55. Find the largest number.

Solution:-

• Let the two numbers be x and y

According to the question:-

• x - y = 5 .... (i)
• x² - y² = 55 .... (ii)

From equation (i)

x - y = 5

=> x = 5 + y .... (iii)

Putting the value of x in equation (ii) from (iii)

= x² - y² = 55

= (5 + y)² - y² = 55

= (5)² + 2 × 5 × y + (y)² - y² = 55

= 25 + 10y + y² - y² = 55

= 25 + 10y = 55

= 10y = 55 - 25

= 10y = 30

$= \sf{y = \dfrac{30}{10}}$

= y = 3

Putting the value of y in equation (iii)

x = y + 5

= x = 3 + 5

= x = 8

∴ The larger number is 8.

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Question:-

Solve the following quadratic equation:-

$\sf{\dfrac{x - 1}{x - 2} - \dfrac{x - 2}{x - 3} = \dfrac{x - 5}{x - 6} - \dfrac{x - 6}{x - 7}}$

Solution:-

$\sf{\dfrac{(x - 1)(x - 3) - (x - 2)^2}{(x - 2)(x - 3)} = \dfrac{(x - 5)(x - 7) - (x - 6)^2}{(x - 6)(x - 7)}}$

$= \sf{\dfrac{[x(x - 3) - 1(x - 3)] - [x^2 - 4x + 4]}{x(x - 3) - 2(x - 3)} = \dfrac{[x(x - 7) - 5(x - 7)] - [x^2 - 12x + 36]}{x(x - 7) - 6(x - 7)}}$

$= \sf{\dfrac{x^2 - 3x - x + 3 - x^2 + 4x - 4}{x^2 - 3x - 2x + 6} = \dfrac{x^2 - 7x - 5x + 35 - x^2 + 12x - 36}{x^2 - 7x - 6x + 42}}$

$= \sf{\dfrac{-4x + 4x - 1}{x^2 - 5x + 6} = \dfrac{-12x + 12x - 1}{x^2 - 13x + 42}}$

$= \sf{\dfrac{-1}{x^2 + x + 6} = \dfrac{-1}{x^2 - 13x + 42}}$

$= \sf{-1(x^2 - 13x + 42) = -1(x^2 - 5x + 6)}$

$= \sf{-x^2 + 13x - 42 = -x^2 + 5x - 6}$

$= \sf{-x^2 + x^2 + 13x - 5x = 42 - 6}$

$= \sf{8x = 36}$

$= \sf{x = \dfrac{36}{8}}$

$= \sf{x = \dfrac{9}{2}}$

$\sf{\therefore\:The\:value\:of\:x\:is\:\dfrac{9}{2}}$

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