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Sia ? 6 years, 4 months ago
We have 2x2 - 7x + 3 = 0
{tex}\implies2( x^2 - {7 \over 2}x + {3\over 2}) = 0{/tex}
{tex}\implies x^2 - {7 \over 2}x + {49 \over 16} = {-3 \over 2} +{ 49 \over 16}{/tex} (Adding 49/16 to both sides)
{tex}\implies x^2 -2 \times x \times {7 \over 4} + ({7 \over 4})^2 = {-24 +49 \over 16}{/tex}
{tex}\implies (x-{7\over4})^2 = {25 \over 16}{/tex}
{tex}\implies x-{7\over 4}= \pm \sqrt({25 \over 16}){/tex}
{tex}\implies x={7\over 4} \pm {5 \over 4}{/tex}
{tex}\implies x={7\over 4} + {5 \over 4}\, and \,x={7\over 4} - {5 \over 4}{/tex}
{tex}\implies x=3\, and \,{1\over 2}{/tex}
{tex}\therefore{/tex}the roots of the given equation are {tex}3{/tex} and {tex}1\over 2{/tex}.
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Ram Kushwah 6 years, 10 months ago
p=a^2b
q=ab^2
than LCM(p,q)=a^2b^2( highest common power of a and b in p and q)
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Jaideep Yadav 6 years, 10 months ago
0Thank You