Question

without solving comment on the nature of the roots x^2-ax -b^2....I know it will be more than zero i.e real and distinct but my question is how to show that by discriminant please dont spam. someone plz answer correctly it's urgent​

Answer 1

Solution:-

We have the given polynomial

$\longrightarrow \sf{ x^2 - ax - b^2}$

Now to have distinct real roots Discriminant (D) should be greater than zero .

$\implies \sf{ b^2 - 4ac > 0 }$

In our polynomial

• a = 1

• b = -a

• c = - b²

$\implies \sf{ D = (-a)^2 - 4.(1)(-b^2) }$

$\implies \sf{ D = a^2 + 4b^2}$

Now as for a real number , square of any real number is greater than or equal to zero and our Discriminant is a sum of squares , So D ≥ 0 .

But we want a condition where D > 0

$\implies \sf { a \neq b \neq 0 }$

So, if the condition is provided we will get real and distinct roots.

If a = b = 0 , then we will get two real roots of same value.

Answer 2

$\sf{\huge{\underline{\boxed{\pink{\mathfrak{ Question}}}}}}$

• find the nature of the roots .

$x^2 - ax - b^2 = 0$

$\sf{\huge{\underline{\boxed{\pink{\mathfrak{ solution}}}}}}$

$\implies \sf x^2 - ax - b^2 = 0$

⠀⠀$\begin{tabular}{|c|c|}\cline{1-2}\sf a &\sf 1 \\\cline{1-2}\sf b &\sf -a\\\cline{1-2}\sf c&\sf -b^2\\\cline{1-2}\end{tabular}$

$\sf{\underline{\boxed{\mathfrak{\purple{ Discriminant = b^2 - 4ac}}}}}$

$\sf\implies { D = ( - a )^2 - 4 \times 1 \times (-b)^2}$

$\sf\implies D = a^2 - 4b^2$

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• since $a^2 + 4b^2$ is not a perfect square therefore roots are irrational
• since D > 0 roots are unequal .
• since highest power is 2 therefore there are 2 roots .

$\sf{\underline{\boxed{\mathfrak{\purple{Equation\: has \: 2 \: unequal \: and \: irrational\:roots}}}}}$

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