Question

1) Find the value of p if the roots of x²+px+12 =0 are in the ratio 1:3 .

2)Find the value of k ,so that the difference of roots x ²- 5x+3(k-1)=0 is 11.

$3) \: solve \: \frac{x - 3}{x} + \frac{x - 1}{x + 2} = \frac{29}{20}$

Answer 1

Heya !!

(1) Let zeroes of the given quadratic polynomial is X and 3X .

P( X ) = X² + PX + 12 = 0

Here,

A = Coefficient of X² = 1

B = Coefficient of X = P

and,

C = Constant term = 12

Therefore,

Product of zeroes = C/A

X + 3X = 12/1

3X² = 12

X² = 12/3

X² = 4

X = ✓2 = 2

First zero = X = 2

and,

Second zero = 3X = 3 × 2 = 6

Sum of zeroes = -B/A

2 + 6 = -P/1

8 = -P

P = -8

(2) X² - 5X + 3 ( K -1 ) = 0

Here,

A = Coefficient of X² = 1

B = Coefficient of X = -5

and,

C = Constant term = 3( K -1 ) = 3K - 3

Therefore,

Sum of zeroes = -B/A

Alpha + Beta = - (-5) / 1

Alpha + Beta = 5 --------(1)

Given that : Difference of two zeroes = 11

So,

Alpha - Beta = 11--------(2)

From equation (1) we get,

Alpha + Beta = 5

Alpha = ( 5 - Beta ) --------(3)

Putting the value of Allah in equation (2)

Alpha - Beta = 11

5 - Beta - Beta = 11

-2Beta = 11 - 5

-2 Beta = 6

Beta = -3

Putting the value of Beta in equation (3)

Alpha = 5 - Beta = 5 - (-3) = 8

------------------------------

Product of zeroes = C/A

-3 × 8 = 3K - 3

-24 = 3K - 3

3K = -24 + 3

K = -21/3

K = -7.

(3) ( X - 3 / X ) + ( X -1 / X + 2 ) = 29/20

=> ( X + 2 ) ( X - 3 ) + ( X - 1 ) ( X ) / ( X ) ( X + 2 ) = 29/20

=> X ( X + 2 )-3 ( X + 2 ) + X² - X / X²+2X =29/20

=> X² + 2X - 3X - 6 + X² - X / X² + 2X = 29/20

=> 2X² - 2X - 6 = 29 ( X² + 2X ) /20

=> 20 ( 2X² - 2X - 6 ) = 29X² + 58X

=> 40X² - 40X - 120 = 29X² + 58X

=> 40X² - 29X² - 40X - 58X - 120 = 0

=> 11X² - 98X - 120 = 0

=> 11X² - 110X + 12X - 120 = 0



=> 11X ( X - 10 ) + 12 ( X - 10 ) = 0




=> ( 11X + 12 ) = 0 OR ( X - 10 ) = 0




=> X = -12/11 or X = 10

Answer 2

Heya!!☻

➢Here's your solution!!!
✧____________________________✧

1) x²+ px +12=0

Ratio - 1:3

Roots are x and 3x

=> x²+ px +12= 0

= a = β = -b/a
= x+3x = -p/1

= 4x= -p/1 .......(1)

= aβ = 12

3x²=12
x²=
x = ±2

In equation ①

8 = -p ︴ +8 = +p
p = -8 ︴ -p = 8

✎__________________________✰

2) Let a and b are the roots of the quadratic equation.

Given quadratic equation is x²-5x+3(k-1) = 0

a-b = 11 -------------(1)

On comparing with ax²+bx²+c

a = 1, b = -5, c = 3(k-1).

sum of a zeros (a+β) = -b/a

(a+β) = -b/a
(a+β) = -(5)/1 = 5
(a+β) = 5-------------(2)

on adding equation 1 and 2.

a-β = 11
a+β =5
______
2a = 16
a = 16/2
a = 8

On putting a = 8 in eq 1

a-β = 11
8 - β = 11
8 - 11 = β
β = -3

Product of zeroes (a•b) = c/a

8×-3 = 3(k-1)/1 [a = 8, β = -3]
-24 = 3(k-1)
-24 = 3k-3
-24+3 = 3k
-21 = 3k
k = -21/3

K = -7

Hence, the value of k = -7

✎________________☻________✰

$\frac{x - 3}{x} + \frac{x - 1}{x - 2} = \frac{29}{20} \\ = \frac{(x + 2)(x - 3) + (x - 1)(x)}{( x \times x + 2)} = \frac{29}{20 } \\ we \: take \: common \\ = \frac{x(x + 2) - 3(x + 2) + {x}^{2} - x }{ {x}^{2} + 2x} = \frac{29}{20} \\ = {2x}^{2} - 2x - 6 = \frac{29( {x }^{2} + 2x)}{20} \\ = 20( {2x}^{2} - 2x - 6) = 29 {x}^{2} + 58x \\ = 40 {x}^{2} - 40x - 120 = 29 {x}^{2} + 58x \\ = 40 {x}^{2} - 29 {x}^{2} - 40x - 58x - 120 = 0 \\ = 11 {x}^{2} - 98x - 120 = 0 \\ = 11 {x}^{2} - 110x + 12x - 120 = 0 \\ = (11x + 12) = 0 \\ = (x - 10) = 0 \\ so \: that \\ x = \frac{ - 12}{11} \: \: \: \: \: \: and \: x = 10$

✎________________________✭
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@vaibhav246