Given that x-root 5 is a …

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Sia ? 6 years, 6 months ago
On factorising the quotient, we get
{tex}x ^ { 2 } - 2 \sqrt { 5 } x - 15 = x ^ { 2 } - 3 \sqrt { 5 } x + \sqrt { 5 }x-15{/tex}
{tex}= x ( x - 3 \sqrt { 5 } ) + \sqrt { 5 } ( x - 3 \sqrt { 5 } ){/tex}
{tex}= ( x + \sqrt { 5 } ) ( x - 3 \sqrt { 5 } ){/tex}
{tex}\therefore\ {/tex}{tex}( x + \sqrt { 5 } ) ( x - 3 \sqrt { 5 } ) = 0{/tex}
{tex}\Rightarrow\ x = - \sqrt { 5 } , 3 \sqrt { 5 }{/tex}
Therefore, all the zeroes are {tex}\sqrt { 5 } , - \sqrt { 5 } \text { and } 3 \sqrt { 5 }{/tex}.
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