Question

please answer the questions 3,4 of challengers

Answer 1

4. x² + px + q = 0
x² + 17x + q = 0

let the roots be a and b.

axb = q/1
30 = q/1
30 = q

original eq. --> x² + 13x + 30 =0
x² + (10+3)x + 30 = O
x² + 10x + 3x + 30 = O
x(x+10) + 3(x+10) = 0
(x+3)(x+10) = 0
x+3=0, x = -3
x+10= 0 , x= -10

Answer 2

Hi,

given, x^2+px+q

the coefficient of x if we take 17 in place of 13.then the roots are -2,-15

x^2+17x+q=0--------(equation1)

alpha=-2 ; beta= -15

alpha+beta= -2-15 = -17

alpha.beta= -2×-15 = 30

then the equation x^2-(alpha+beta)x+alpha.beta

=>x^2-(-17)x+30

=>x^2+17x+30------------(equation2)

compare equation (1)&(2)
:.q=30

The real equation x^2+13x+30=0

x^2+10x+3x+30=0

x (x+10)+3 (x+10)=0

(x+10)(x+3)=0

(x+10)=0 (or) (x+3)=0

x= -10 (or) x= -3

:. the real roots are -10,-3

(3rd answer I given in pic)

hope it helps.....