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(C2 ab)x2-(a2 -bc)x+b2 ac=0 has equal roots prove that a=0

Posted by Tarushi Khattar 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

We have, {tex}(c^2-ab)x^2-2(a^2-bc)x+b^2-ac=0{/tex}

Here, {tex}A=(c^2-ab), B=-2(a^2-bc),C=b^2-ac{/tex}

Now, {tex}B^2-4AC =0{/tex}

{tex}4(a^2-bc)^2-4(c^2-ab)(b^2-ac)=0{/tex}

{tex}4(a^2-bc)^2=4(c^2-ab)(b^2-ac){/tex}

{tex}a^4-2a^2bc+b^2c^2=b^2c^2-ac^3-ab^3+a^2bc{/tex}

{tex}a^4-2a^2bc+ac^3+ab^3-a^2bc=0{/tex}

{tex}a^4-3a^2bc+ac^3+ab^3=0{/tex}

{tex}a(a^3-3abc+c^3+b^3)=0{/tex}

{tex}a=0 \ or \ a^3-3abc+c^3+b^3=0{/tex}

{tex}a=0\ or\ a^3+b^3+c^3=3abc{/tex}

Hence proved

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CBSE > Class 10 > Mathematics