Question

a, b are zeros of the polynomials x2-bx+c find the value of a, if 3a+2b=20​

Answer 1

Step-by-step explanation:

As a and b are zeroes of the polynomial

x^2-bx+c. Hence substitute x=>a and x=>b in this polynomial and equate it to zero.

${a}^{2} - ab + c = 0 \: and \: {b}^{2} - {b}^{2} + c = 0$

$= > c = 0$

Then

${a}^{2} - ab = 0 = > a(a - b) = 0 = >$

$a = 0 \: or \: a = b$

If a=0, 2b=20=>b=10 So the polynomial becomes

${x}^{2} - 10x = 0 = > x(x - 10) = 0 = >$

$x = 0 \: or \: 10$

If a=b,3a+2a=20=>5a=20=>a=4 and b=4.

$so \: {x}^{2} - 4x = 0 = > x(x - 4) = 0 = >$

$x = 0 \: or \: 4$

Hence a can't be equal to b Hence a=0 and b=10. Hope it helps you.