Question

plz answer this question

Answer 1

Solution:

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Given:

ax² + bx + c = 0,

α : β = m : n ,

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Prove that :

=> mnb² = ac(m + n)²

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As we know,


value of x = $\frac{-b + \sqrt{b^2 - 4ac} }{2a} (or) \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

α = $\frac{-b + \sqrt{b^2 - 4ac} }{2a}$

β = $\frac{-b - \sqrt{b^2 - 4ac} }{2a}$

∴  m : n = $\frac{-b + \sqrt{b^2 - 4ac} }{2a} : \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

Proof:

LHS = mnb²

=> $( \frac{-b + \sqrt{b^2 - 4ac} }{2a}) ( \frac{-b - \sqrt{b^2 - 4ac} }{2a})b^2$

=> $( \frac{-b}{2a} +\frac{ \sqrt{b^2 - 4ac} }{2a})( \frac{-b}{2a} - \frac{- \sqrt{b^2 - 4ac} }{2a})b^2$

We know that,

(a + b)(a - b) = a² - b²,

here,

we take

a = $\frac{-b}{2a}$

&

b = $\frac{ \sqrt{b^2 - 4ac} }{2a}$

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=> $(( \frac{-b}{2a})^2 - ( \frac{ \sqrt{b^2 - 4ac} }{2a} )^2 ) b^2$

=> $( \frac{b^2}{4a^2} - \frac{b^2 - 4ac}{4a^2} ) b^2$

=> $( \frac{b^2 - b^2 + 4ac}{4a^2} )b^2$

=> $( \frac{4ac}{4a^2})b^2$

=> $\frac{c}{a}(b^2)$

=> $\frac{b^2c}{a}$ .........................

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RHS = ac(m + n)²

=> $ac( \frac{-b + \sqrt{b^2 - 4ac} }{2a} + \frac{-b - \sqrt{b^2 4ac} }{2a} )^2$

=> $ac( \frac{-b + \sqrt{b^2 - 4ac } -b - \sqrt{b^2 - 4ac} }{2a} )^2$

=> $ac( \frac{-2b}{2a})^2$

=>$ac( \frac{-b}{a} )$

=> $ac( \frac{b^2}{a^2} )$

=> $\frac{b^2 c}{a}$  .................

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LHS = RHS,

Hence proved,.

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