Question

If ✓(13-a√10) = √3+√5 then a=?​

Answer 1

Answer:

$\sqrt{13 - a \sqrt{10} } = \sqrt{3 } + \sqrt{5} \\ \\ \\ = > \sqrt{13} - 10 \sqrt{a} = \sqrt{3} + \sqrt{5} \\ \\ \\ = > \sqrt{13} - \sqrt{3} - \sqrt{5} = 10 \sqrt{a} \\ \\ \\ \\ \\ = > 3.60 - 1.732 - 2.23 = 10 \sqrt{a} \\ \\ \\ = > 3.605 - 3.96 = 10 \sqrt{a} \\ \\ \\ = > - 0.3 = 10 \sqrt{a} \\ \\ \\ \\ = > \sqrt{a} = \frac{ - 0.3}{10} \\ \\ = > \sqrt{a} = \frac{ - 3}{100} \\ \\ = > a = - \frac{ \sqrt{3 } }{10}$

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Answer 2

Answer:

$\ a = \frac{(\sqrt{5}-2*\sqrt{3} )}\sqrt{2}}$

Step-by-step explanation:

$\sqrt{13-a\sqrt{10} } =\sqrt{3} -\sqrt{5}$

⇒$\ 13-a\sqrt{10} } =(\sqrt{3} -\sqrt{5})^{2}$

using $(a+b)^{2} =a^{2} +b^{2} +2ab$

⇒$\ 13-a\sqrt{10} } =( 3 + 5 +2*\sqrt{3} *\sqrt{5})$

⇒$\ -a\sqrt{10} } = -5 +2*\sqrt{3} *\sqrt{5}$

⇒$\ a = \frac{5 -2*\sqrt{3} *\sqrt{5} }\sqrt{10}}$

⇒$\ a = \frac{(\sqrt{5}*\sqrt{5}-2*\sqrt{3} *\sqrt{5} )}\sqrt{5*2}}$

⇒$\ a = \frac{(\sqrt{5}-2*\sqrt{3} )}\sqrt{2}}$