Question

find the zeroes of polynomial hsquare-x-12 and verify relationship between the zeroes and cofficeint​

Answer 1

$\texttt{\textsf{\large{\underline{Solution}:}}}$

Given equation,

$\tt \implies {x}^{2} - x - 12 = 0$

By splitting the middle term, we get,

$\tt \implies {x}^{2} - 4x + 3x- 12 = 0$

$\tt \implies x(x - 4) + 3(x- 4) = 0$

$\tt \implies (x+ 3)(x- 4) = 0$

By zero product rule,

$\begin{cases} \tt x + 3 = 0 \\ \tt x - 4 = 0 \end{cases}$

Therefore,

$\tt \implies x = 4, - 3$

$\texttt{\textsf{\large{\underline{Verification}:}}}$

Comparing the given equation with ax² + bx + c = 0, we get,

$\begin{cases} \tt a = 1 \\ \tt b = - 1 \\ \tt c = - 12\end{cases}$

We know that,

→ Sum of roots = -b/a

→ Product of roots = c/a

Here,

→ Sum of roots = 4 - 3 = 1

→ Sum of roots = -b/a = -(-1)/1 = 1

Hence Verified!

Again,

→ Product of roots = 4 × (-3) = -12

→ Product of roots = c/a = -12/1 = -122

Hence Verified!n

$\texttt{\textsf{\large{\underline{Know More}:}}}$

Discriminant: The discriminant of any equation tells us about the nature of roots.

The general form of a quadratic equation is,

$\tt\implies ax^{2} + bx+c=0$

Discriminant is calculated by using the formula given below,

$\tt\implies D=b^{2}-4ac$

When D > 0: Roots are real and distinct.

When D < 0: Roots are imaginary.

When D = 0: Roots are real and equal.

Answer 2

Answer:

First step : Finding zeroes

Given,

${x}^{2} - x - 12\\ = {x}^{2} - 4x + 3x - 12 \\ = x(x - 4) + 3(x - 4) \\ = (x - 4)(x + 3) \\ \\ \bf \: for \: zeroes \: \: \: either \\ \: x - 4 = 0 \: \: \: \therefore \: \pink{x = 4 }\\ \bf \: or \\ x + 3 = 0 \: \: \: \: \therefore \: \pink{ x = - 3} \\ \\ \bf \: so \: zeroes \: \: are \: \: \green{ \boxed{x = 4 \: \: or \: - 3}}$

Second step :Verify the relationship between the zeros and the coefficient.

$\sf \: if \: \alpha \: \: and \: \beta \: are \: the \: zeroes \: of \: the \: \\ \sf polynomial \: \: \: a {x}^{2} + bx + c \: \: \: \: then \: \\ \\ \\ \sf \: sum \: of \: zeroes \: \: \: \: \alpha + \beta = - \frac{b}{a} \: \: \: \: \: \: and \\ \\ \sf \: \: product \: of \: zeroes \: \: \alpha \beta = \frac{c}{a}$

Here,

Zeroes of the polynomial are 4 and - 3

So,

sum of zeroes is = 4+(-3) = 4-3 = 1

and

product of zeroes is = 4x(-3) = -12

again,

From the polynomial

$\alpha + \beta = - \frac{ - 1}{1} = 1 \\ \\ \alpha \beta = \frac{ - 12}{1} = - 12$

Hence verified.