Question

if a,b are zeros of the quadratic polynomial f(x) = 2x^2+11x+5 , find (a). a^4+b^4 (b). 1/a+1/b-2ab
its unjent please ​

Answer 1

Answer:

• (a) a⁴ + b⁴ = 625
• (b) 1/a + 1/b - 2ab = -36/5

Explanation:

Given polynomial,

⇒ 2x² + 11x + 5

On comparing with the general form of equation ax² + bx + c

• a = 2
• b = 11
• c = 5

Using quadratic formula,

$\implies \boldsymbol{x \: = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }$

Substituting the values,

${\implies \boldsymbol{x = \dfrac{ - 11 \pm \sqrt{ {(11)}^{2} - 4(2)(5)} }{2(2)} }}$

$\implies \boldsymbol{x = \dfrac{ - 11 \pm \sqrt{121 - 40} }{4} }$

$\implies \boldsymbol{x = \dfrac{ - 11 \pm \sqrt{81} }{4} }$

$\implies \boldsymbol{x = \dfrac{ - 11 \pm9}{4} }$

$\implies \boldsymbol{x = \dfrac{ - 20}{4} \: { \rm \: or} \: \dfrac{ - 2}{4} }$

$\implies \boldsymbol{x = \dfrac{ - 1}{2} \: { \rm \: of} \: - 5 }$

$\rm \therefore \: zeros \: of \: the \: polynomial \: are \: \dfrac{ - 1}{2} \: and \: - 5$

Now, Solving (a) :- a⁴ + b⁴

$= { \bigg( \dfrac{ - 1}{2} \bigg)}^{4} + \big( - 5 \big)^{4}$

$= \dfrac{1}{16} + 625$

$= \dfrac{10000}{16}$

$= {625}$

$\boldsymbol{\therefore {a}^{4} + {b}^{4} = 625}$

And also, for (b) 1/a + 1/b - 2ab

$= \dfrac{1}{a} + \dfrac{1}{b} + 2ab$

$= \dfrac{1}{ \frac{ - 1}{2} } + \dfrac{1}{ - 5} - 2 \bigg( \dfrac{ - 1}{2} \bigg)( - 5)$

$= - 2 + \dfrac{ - 1}{5} - 5$

$= \dfrac{ - 10 - 1 - 25}{5}$

$= \dfrac{ - 36}{5}$

$\therefore \boldsymbol{ \dfrac{1}{a} + \dfrac{1}{b} - 2ab = \dfrac{ - 36}{5}}$

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