Question

Find the zeroes of the Xsquare2+5x-6 diagonal polynomial and check the correctness of the relationship between the zeros and the coefficients.

Answer 1

AnswEr:-

Zeroes of polynomial = -6 & 1

$\rule{200}{1}$

Given polynomial:-

☛ x² + 5x - 6

Now by factorization method:-

$\dashrightarrow\sf x^2 + 5x - 6 \\\\\dashrightarrow\sf x^2 + 6x - x - 6 \\\\\dashrightarrow\sf x(x + 6) -1(x + 6) \\\\\dashrightarrow\sf (x + 6)(x - 1) \\\\\dashrightarrow{\underline{\boxed{\textsf{\textbf{x = - 6 \: or \: x = 1 }}}}}$

$\therefore\underline{\textsf{Zeroes of polynomial = {\textbf{-6 \& 1 }}}}$

Now checking the correction of the relationship between the zeros and coefficients.

Here,

• Coefficient of x²(a) = 1
• Coefficient of x(b) = 5
• Constant term(c) = -6

Relationship 1:-

☛ Sum of zeroes = -b/a

↠ -6 + 1 = -5/1

↠ -5 = -5 [Verified!]

Relationship 2:-

☛ Product of zeroes = c/a

↠ -6 × 1 = -6/1

↠ -6 = -6 [Verified!]

Therefore,

Zeroes of the polynomial are -6 & 1.

Answer 2

Correct Question

Find the zeroes of the polynomiaX²+5x-6 and check the relationship between the zeros and the coefficients.

Solution

px=X²+5x-6

Factorising it for zeroes

$\implies x^2+5x-6$

$\implies$X²+6x-x-6

$\implies$X(X+6)-1(X+6)

$\implies$ (X-1)(X+6)

Then, zeroes are

X=(1,-6)

____________________

Now

Let the zeroes of the polynomial be $\alpha and \beta$

Then

$\alpha + \beta$=1-6= -5

$\frac{-5}{1}$=$\frac{-(cofficient\:of\:x)}{Cofficient\:of\:x^2}$

For

$\alpha \times \beta$=-6

$\frac{-5}{1}$ =$\frac{(constant\:term)}{Cofficient\:of\:x^2}$

_____________________

Here,the relation between Coefficient and zeros