Question

CLASS X MATHS...
1.If α and β are zeroes of the polynomial x²-2x-15 then form a quadratic polynomial whose zeroes 2α and 2β.
2.If the sum of the zeroes of the polynomial f(t)=kt²+2t+3k is equal to the product then the value of k is________?
3.Find the value of a for which (x-a) is a factor of f(x) =-x³+ax²+3x+9.
4.If x-a/b+a + x-b/c+a + x-c/a+b = 3 then the value of x is______?

Answer 1

1)      α+β = 2            αβ = -15
       2α+ 2β = 4        4αβ = - 60
             x² - 4 x - 60
2)  - 2 / k =  3k/k            =>  k = -2/3

3)   f(a) = - a³ + a a² + 3a + 9 =  3 a + 9 = 0    => a = -3
              
4)  
$\frac{x-a}{b+c}+\frac{x-b}{c+a}+\frac{x-c}{a+b}=3\\\\\frac{x-a}{b+c}+\frac{x-b}{c+a}=3-\frac{x-c}{a+b}\\\\\frac{xc+ax-ac-a^2+xb+xc-bc-b^2}{(b+c)(a+c)}=\frac{3a+3b-x+c}{a+b}\\\\\frac{x(a+b+2c)-(a^2+b^2+ac+bc)}{ab+ac+bc+c^2}=\frac{3a+3b+c-x}{a+b}$

[tex](a+b)*[x(a+b+2c)-(a^2+b^2+ac+bc)]\\.\ \ \ \ \ = (ab+ac+bc+c^2) * (3a+3b+c-x)\\\\x[(a+b)(a+b+2c)+(ab+ac+bc+c^2)]=\\.\ \ \ \ (ab+ac+bc+c^2) * (3a+3b+c)+(a+b)(a^2+b^2+ac+bc)\\\\x[a^2+3ab+b^2+3ac+3bc+c^2][/tex]

[tex].\ \ \ \ \ =(3a^2b+3a^2c+7abc+4ac^2+3ab^2+3b^2c+4bc^2+c^3)\\.\ \ \ \ \ \ \ \ +a^3+ab^2+a^2c+abc+a^2b+b^3+abc+b^2c\\\\x[a^2+3ab+b^2+3ac+3bc+c^2]\\.\ \ \ \ =a^3+b^3+c^3+9abc+4a^2b+4a^2c+4ac^2+4ab^2+4b^2c+4bc^2\\\\x=\frac{a^3+b^3+c^3+9abc+4a^2b+4a^2c+4ac^2+4ab^2+4b^2c+4bc^2}{a^2+3ab+b^2+3ac+3bc+c^2}[/tex]

$x=\frac{(a+b+c)^3+3abc+ab(a+b)+ac(a+c)+bc(b+c)}{(a+b+c)^2+(ab+bc+ca)}$

Answer 2

Answer:

$\huge{Your\:Answer}$

1.Sol: ➡️Given that α and β are the zeroes of the polynomial x2 - 2x

-(x-a-b-c)*k 15 then α + β = 2 and αβ  = -15 If 2α, 2β are zeros of the

quadratic polynomial then the equation is  x2 - 2(α + β)x + 4αβ =0 then Sum of roots = 2(α + β) = 4              Product of roots = 4αβ

 = -60 Now the polynomial equation is x2 - 4x - 60 =0.

2.➡️Given quadratic polynomial f(t) = kt2  + 2t +3k α,β are the roots of the quadratic polynomial then α+β = -2/k and αβ = 3k/k According to the condition α+β = αβ -2/k = 3∴ k = -2/3.

3.➡️Given,

x - a is a factor of 

{x}^{3} - ax {}^{2} + 2x + a - 1x3−ax2+2x+a−1 

we know,

x - a = 0

x = 0+a = a

Substitung the value of x as a,

we have,

(a)^3 - a(a)^2 + 2(a) + a - 1 =0

a^3 - a^3 + 2a + a - 1=0

3a - 1=0

3a =1

a = 1/3

Thus

a = 1/3

4.➡️(x-a-b-c)*K=0

Note -k is some random expression in a , b and c.

This implies =

x-a-b-c=c

x=a+b+c