23. If aB are zeroes of the quadratic polynomial x-7x+10, find the val

1.	23. If aB are zeroes of the quadratic polynomial x-7x+10, find the value of
Answer» x^2-7x+10=0; x^2-5x-2x+10=0; x(x-5)-2( x-5)=0; ( x-5)(x-2); x=2, 5 X²+7x+10=0 x²+2x+5x+10=0 x(x+2)+5(x+2)=0 (x+2)(x+5) = 0 x+2 = 0 ; x = -2 x+5 = 0 ; x = -5 Relationship between the zeroes and coefficients :- Sum of zeroes = -2+(-5) = -2-5 = -7/1 = -x coefficient /x² coefficient Product of zeroes = (-2)(-5) = 10/1 = constant/x² coefficient Hope it helps x^2+7x+10=0; x^2+2x+5x+10=0; x( x+2)+5( x+2)=0 ( x+5)( x+2); x=2, 5 ( or) -2, -5 let alpha=p bita= qx^2-7x+10sum of zeros = -b/ap+q= 7product of zeros= c/apq = 10p^2+q^2= (p+q)^2-2pq = (7)^2-2×10 = 49-20= 29 x=2,5 correct answer is 2,5

23. If aB are zeroes of the quadratic polynomial x-7x+10, find the value of

Answer»

x^2-7x+10=0; x^2-5x-2x+10=0; x(x-5)-2( x-5)=0; ( x-5)(x-2); x=2, 5

X²+7x+10=0

x²+2x+5x+10=0

x(x+2)+5(x+2)=0

(x+2)(x+5) = 0

x+2 = 0 ; x = -2

x+5 = 0 ; x = -5

Relationship between the zeroes and coefficients :-

Sum of zeroes = -2+(-5) = -2-5 = -7/1 = -x coefficient /x² coefficient

Product of zeroes = (-2)(-5) = 10/1 = constant/x² coefficient

Hope it helps

x^2+7x+10=0; x^2+2x+5x+10=0; x( x+2)+5( x+2)=0 ( x+5)( x+2); x=2, 5 ( or) -2, -5

let alpha=p bita= qx^2-7x+10sum of zeros = -b/ap+q= 7product of zeros= c/apq = 10p^2+q^2= (p+q)^2-2pq = (7)^2-2×10 = 49-20= 29

x=2,5

correct answer is 2,5