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

Step-by-step explanation:

1)Ans:-

$\alpha + \beta =\frac{2}{1} = 2$

$\alpha\beta =\frac{-15}{1} = -15$

$Form \: of \: that \: quadratic \: polynomial-$

$Whose \: zeroes \: are \: 2\alpha \: and \: 2\beta$

$x²-(2\alpha+2\beta)x+(2\alpha)(2\beta)=0$

$x² -2(\alpha + \beta)x + 4\alpha\beta = 0$

$x² - 2(2)x + 4(-15) = 0$

$x² - 4x - 60 = 0$

2)Ans:-

$\alpha + \beta = \alpha\beta$

$\frac{-2}{k} = \frac{3k}{k}$

$\frac{-2}{k} = 3$

$k =\frac{-2}{3}$

3)Ans:-

$(x - a) = 0$

$x = a$

$f(a) = 0$

$-(a)³ + a(a)² + 3a + 9 = 0$

$-a³ + a³ + 3a + 9 = 0$

$3a + 9 = 0$

$3a = -9$

$a = -3$