Question

Pls answer question 21 vi, vii, viii
Pls answer fast and correct will mark you as brainliest.​

Answer 1

Good question I am answering the 4th question

Answer 2

Given :-

• f(x)= ax² + bx + c
• α and β are zeroes of f(x)

To find :-

$\sf{(1)\bullet \frac{1}{a\alpha+b}+\frac{1}{a\beta+b} }$

$\sf{(2)\bullet \frac{\beta}{a\alpha + b} + \frac{\alpha}{a\alpha+b}}$

$\sf{(3)\bullet \:\: a\left( \frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\right)+b\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)}$

Solution :-

In the given quadratic polynomial f(x)

$\boxed{\sf{\alpha+\beta=\frac{-b}{a}}\;\;...eqn(1)}$

and

$\boxed{\sf{\alpha\beta=\frac{c}{a}}\;\;...eqn(2)}$

$\rule{300}3$

Solution (1)

$\implies\sf{\frac{1}{a\alpha+b}+\frac{1}{a\beta+b}}$

taking LCM

$\implies\sf{\frac{a\beta+b+a\alpha+b}{a^2\alpha\beta+\alpha\;ab+\beta\;ab+b^2}}\\\\\implies \sf{\frac{a(\alpha+\beta)+2b}{a^2\alpha\beta+ab(\alpha+\beta)+b^2}}$

using eqn (1) and (2)

$\implies\sf{\frac{a(\frac{-b}{a})+2b}{a^2(\frac{c}{a})+ab(\frac{-b}{a})+b^2} }\\\\\boxed{\blue{\implies\sf{\frac{b}{ac}}}}$

$\rule{300}3$

Solution (2)

$\implies\sf{\frac{\beta}{a\alpha+b}+\frac{\alpha}{a\beta+b}}$

Taking LCM

$\implies\sf{\frac{\beta(a\beta+b)+\alpha(a\alpha+b)}{a^2\alpha\beta+ab\;\alpha+ab\;\beta+b^2} }\\\\\implies\sf{\frac{a\beta^2+b\beta+a\alpha^2+b\alpha}{a^2\alpha\beta+ab(\alpha+\beta)+b^2} }\\\\\implies\sf{\frac{a(\alpha^2+\beta^2)+b(\alpha+\beta)}{a^2\alpha\beta+ab(\alpha+\beta)+b^2} }\\\\\implies\sf{\frac{a((\alpha+\beta)^2-2\alpha\beta)+b(\alpha+\beta)}{a^2\alpha\beta+ab(\alpha+\beta)+b^2} }$

using eqn (1) and (2)

$\implies\sf{\frac{a((\frac{-b}{a})^2-2(\frac{c}{a}))+b(\frac{-b}{a})}{a^2(\frac{c}{a})+ab(\frac{-b}{a})+b^2} }\\\\\boxed{\blue{\implies\sf{\frac{-2}{a}}}}$

$\rule{300}3$

Solution (3)

$\implies\sf{a\left(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\right)+b\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)}\\\\\implies\sf{a\left(\frac{\alpha^3+\beta^3}{\alpha\beta}\right)+b\left(\frac{\alpha^2+\beta^2}{\alpha\beta}\right )}\\\\\implies\sf{a\left(\frac{(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)}{\alpha\beta}\right)+b\left(\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}\right)}$

using eqn (1) and (2)

$\implies\sf{a\left(\frac{(\frac{-b}{a})^3-3(\frac{c}{a})(\frac{-b}{a})}{\frac{c}{a}}\right)+b\left(\frac{(\frac{-b}{a})^2-2(\frac{c}{a})}{\frac{c}{a}}\right)}\\\\\boxed{\blue{\implies\sf{b}}}$

$\rule{300}3$