Question

for what values of a quadratic equation x^2-ax+1=0 does not have real roots?

Answer 1

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Answer 2

Answer:

if  the value of 'a' is greater then -2 and less then 2   than $x^2 -ax + 1= 0$ does not have real roots .

Step-by-step explanation:

Explanation:

Given , a quadratic equation  $x^2 -ax + 1= 0$

If the discriminant D is greater then 0  , than roots are real and unique.

And if the discriminant value is small then 0 , than the roots are unreal .

Where D = $b^2 - 4ac$ .

Step 1:

From the question we have $x^2 -ax + 1= 0$

a = 1 , b = -a and c = 1

Therefore , for unreal roots ,

D< 0

⇒$b^2 - 4ac$ < 0

⇒$(-a)^{2}$ - 4 × 1× 1 <0

⇒$a^2$ - 4< 0 ⇒ $a^2$ < 4

⇒ a < $\frac{+}{}$2

Final answer:

Hence , if a is greater then -2 and less then 2   than $x^2 -ax + 1= 0$ does not have real roots .

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