Question

Find the zeros of the quadratic polynomial p(x)=s(x^+1)-x(a^+1) and verify the relationship between the zeros and the coefficients.​

Answer 1

Correct Statement is

• Find the zeros of the quadratic polynomial p(x)=a(x^2+1)-x(a^2+1) and verify the relationship between the zeros and the coefficients.

Answer

Given:-

$\tt \: The \:\: polynomial \: p(x) \: = \: a( {x}^{2} + 1) - x( {a}^{2} +1 )$

To find :-

• zeros of the quadratic polynomial
• verify the relationship between the zeros and the coefficients.

Solution:-

$\tt \: p(x) \: = a( {x}^{2} + 1) - x( {a}^{2} + 1)$

$\tt \: p(x) \: = {ax}^{2} + a - {xa}^{2} - x$

$\tt \: p(x) \: = ( {ax}^{2} - {xa}^{2} ) + (a - x)$

$\tt \: p(x) \: = ax(x - a) - 1(x - a)$

$\tt \: p(x) \: = (ax - 1)(x - a)$

Hence, zeros of p(x) are

$:  \implies \boxed{\purple{\bf \: x \: = \: a \: \: or \: \: x \: = \: \dfrac{1}{a} }}$

Now,

$\red{\tt \: Sum \: of \: zeros \: = a \: + \: \dfrac{1}{a} = \dfrac{ {a}^{2} + 1 }{a}} - - (i)$

$\boxed{\red{ \tt \: Product \: of \: zeros \: = a \times \dfrac{1}{a} = 1}} - - (ii)$

Now, p(x) can be rewritten in standard form as

$\tt \: p(x) \: = a( {x}^{2} + 1) - x( {a}^{2} + 1)$

$\tt \: p(x) \: = {ax}^{2} + a - {xa}^{2} - x$

$\tt \: p(x) \: = {ax}^{2} - x( {a}^{2} + 1) + a$

Now, Consider

$\red{ \tt \: - \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} } = \dfrac{ {a}^{2} + 1 }{a}} - - (iii)$

$\red{ \tt \: \dfrac{constant \: term}{coefficient \: of \: {x}^{2} } = \dfrac{a}{a} = 1} - - (iv)$

Now, from (i), (ii), (iii), (iv), we concluded that

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\large{\boxed{\boxed{\bf{Hence, verified}}}}$