Question

Write the set of values of a for which x2+ax-2=0 has real roots

Answer 1

Answer:

Step-by-step explanation:

by the rule of shreedharacharya

roots are

(-a+√a²+8)/2 & (-a-√a²+8)/2

Answer 2

Answer:

$If \: a > ±\sqrt{-8}\: then \\</p><p>the \: equation \: has \: real \: roots$

Step-by-step explanation:

$Compare\: given \\\: quadratic\: equation \: x^{2}+ax-2=0 \: with \\ Ax^{2}+Bx+C=0 \: we\:get$

$A=1 ,\: B= a \: C = -2$

$Discreminant (D) > 0$

\*Given Roots are real *\

$\implies B^{2}-4AC >0$

$\implies a^{2}-4\times 1\times (-2)>0$

$\implies a^{2}+8>0$

$\implies a^{2}> -8$

$\implies a > ±\sqrt{-8}$

Therefore,.

$If \: a > ±\sqrt{-8}\: then \\</p><p>the \: equation \: has \: real \: roots$

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