Question

If P (x) = ax²+bx+c and a+b+c =0, than one zero is ?​

Answer 1

$\mathfrak{\huge\underline{Answer:}}$

x=1 is one of the zero of p(x) .

Step-by-step explanation:

Given p(x)= ax²+bx+c

The above mentioned equation is in the quadratic form

a+b+c =0

Consider the given eqn

p(x)= ax²+bx+c

The above mentioned equation is the general form of the polynomial where a, b, c are variables and x is a constant

Put x=1 in the above equation we get

p(1)=a+b+c

But we know that a+b+c = 0

Therefore p(1) =0

So when x =1 p(1) =0

Therefore x = 1 is the root of p(x)

x = 1 is one of the zero of p(x)

Here p represents the polynomial

$\huge\fbox\red{HOPE\:IT\: HELPS\: YOU}$

Answer 2

$\huge \underline \colorbox{red}{given}$

$= > a + c = b$

$\huge \colorbox{green}{we \: have}$

$= > p(x) = ax^{2} + bx + c \\ \\ = > p(x) = {ax}^{2} + (a + c)x + c \\ \\ = > p(x) = {ax}^{2} + ax + cx + c \\ \\ = > p(x) = ax(x + 1)(ax + c) \\ \\ = > p(x) = (x + 1)(ax + c)$

==》Then by making p(x)=0, we get :-

$x + 1 = 0 \colorbox{red}{and} \: ax + c = 0 \\ \\ = = > x = - 1 \: \: \: \colorbox{cyan}{and} \: \: \: x = \frac{ - c}{a}$

$\huge\bf \underline \colorbox{cyan}{alternate \: method}$

Given

$\ = > a + c = b$

Now , we have

$= > p(x) = {ax}^{2} + bx + c$

$\bf\underline{let \: \: \alpha \: and \: b \: \: are \ \: : the \: \: zeros \: \: of \: \: this \: \: polynomial \: \:p(x) = {ax}^{2} + bx + c}$

we know that :-

$\alpha + \beta = \frac{ - \beta }{a} \colorbox{red}{and} \: \alpha b = \frac{ - c}{a} \\ \\ \bf \for \: finding \: ( \alpha - \beta ) \\ \\ = > ( \alpha - \beta ) ^{2} = { \alpha }^{2} + { \beta }^{2} - 2 \alpha \beta \\ \\ = > ( \alpha - \beta )^{2} = ( { \alpha }^{2} + { \beta }^{2}) - 2 \alpha \beta - 2 \alpha \beta \\ \\ = > ( \alpha - \beta )^{2} = ( \alpha + \beta )^{2} - 4 \alpha \beta$

Substituting the values on R.H.S:--

$= > ( \alpha - \beta )^{2} = ({ \frac{ - b}{a}) }^{2} -4( \frac{c}{a} ) \\ \\ = > {( \alpha - \beta )}^{2} = \frac{ {b}^{2} }{a} - \frac{4c}{a} \\ \\ = > {( \alpha - \beta )}^{2} = \frac{ {b}^{2} }{ {a}^{2} } - \frac{4c}{a} \\ \\ = > {( \alpha - \beta )}^{2} = \frac{ {b } ^{2 -}4ac }{ {a}^{2} } \\ \\ = > {( \alpha - \beta )}^{2} = \frac{(a + c)^{2} - 40c }{ {a}^{2} } \\ = > ( \alpha - \beta )^{2} = \frac{ {(a - c)}^{2} }{ {a}^{2} } \\ \\ = > \alpha - \beta = \frac{(a - c)}{a}$

$\bf \then \: adding \: \: \alpha + \beta = \frac{ - b}{a}$

$\alpha - \beta = \frac{a - c}{a}$

$= > \alpha + \beta + \alpha - \beta = \frac{ - b}{a} + \frac{a - c}{a} \\ \\ = > 2 \alpha = \frac{ - (a + c)}{a} + \frac{a - c}{a} \\ \\ = > 2 \alpha = \frac{ - a - c + a + c}{ \alpha } \\ \\ = > 2 \alpha = \frac{ - 2c}{a} \\ \\ = > \alpha = \frac{ - c}{a}$

$\alpha \beta = \frac{c}{a } \\ \\ = > \frac{ - c}{a} \times \beta = \frac{c}{a} \\ \\ = > \beta = - 1$

Thus the zeroes of the polynomial p(x)= ax²+bx+c are :-

$\huge \bf\frac{ - c}{a} \: \: \colorbox{cyan}{and} \: - 1$