Question

If α and β are the zeroes of the quadratic polynomial f(x) = ax2 + bx + c, then the value pls give proof of this answer​

Answer 1

Answer:

True

Step-by-step explanation:

let roots be x, y as there is no symbol for alpha and beta on the keyboard

then,

prove x²-y² = (-b/a){√(b²-4c)}/a

we know

xy=c/a. ...1

x + y = -b/a. ....2

(x+y)² = (-b/a)²

x²+y²+2xy = b²/a²

subtracting 4xy on both sides

by using xy=c/a

x²+y²-2xy = b²/a²-4*c/a

(x-y)² = ( b²-4ac)/a²

x-y = {√(b² - 4ac)}/a. ...3

then, by 2 and 3

x-y = {√(b² - 4ac)}/a

x + y = -b/a

we know that (x+y)(x-y)= x²-y²

so, multiplying 2*3

x²-y²= (-b/a){√(b²-4ac)}/a

so, the answer is true

Answer 2

Step-by-step explanation:

We know that relation between zeros and coefficient of quadratic equation is given by

Quadratic polynomial

$a {x}^{2} + bx + c = 0 \\ \\ zeros = \alpha \: and \: \beta \\ \\ \alpha + \beta = \frac{ - b}{a} \: \: ...eq1 \\ \\ \alpha \beta = \frac{c}{a}$

Now,to find the value of

$\boxed{{( \alpha - \beta )}^{2} = ( { \alpha + \beta )}^{2} - 4 \alpha \beta } \\ \\ {(\alpha - \beta )}^{2} \: = \frac{ {b}^{2} }{ {a}^{2} } - \frac{4c}{a} \\ \\ {(\alpha - \beta )}^{2} \: = \frac{ {b}^{2} - 4ac}{ {a}^{2} } \\ \\taking \: square \: root \: both \: sides \\ \\ ( \alpha - \beta ) = \: \sqrt{ \frac{ {b}^{2} - 4ac}{ {a}^{2} } } \: \:... eq2$

Multiply eq1 and eq2,because

$(x - y)(x + y) = {x}^{2} - {y}^{2} \\ \\ so \\ \\ ( \alpha + \beta )( \alpha - \beta ) = { \alpha }^{2} - { \beta }^{2} \\ \\ { \alpha }^{2} - { \beta }^{2} = \bigg ( \frac{ - b}{a} \bigg)\bigg(\sqrt{ \frac{ {b}^{2} - 4ac}{ {a}^{2} } } \bigg) \\ \\ { \alpha }^{2} - { \beta }^{2} = \bigg( \frac{ - b}{ {a}} \sqrt{ \frac{ {b}^{2} - 4ac}{ {a}^{2} } } \bigg)$

So,given statement is True.

Hope it helps you.