Question

Determine the nature of the root of the quadercatoric equation
2x²-2x+9=0​

Answer 1

ANSWER:

No real roots

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EXPLANATION:

Here this is a question from quadratic equation, where we have to find the nature of the roots of the given quadratic equation. To find the nature of the roots we have to apply formula of nature of roots and nature of the roots can be expressed as follows:

$\boxed{\boxed{\begin{array}{c |c }\bf{Discriminant}&\bf{Nature}\\\\\sf{+ve}&\sf{2\: distinct \:roots}\\\\\sf{-ve}&\sf{No\:real\:roots}\\\\ \sf{0 }&\sf{Two\: equal\:roots}\end{array}}}$

So let's start!

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Given quadratic equation:

2x²-2x+9=0

Here value of:

→ Coefficient of x², a=2

→ Coefficient of x, b=-2

→ Constant term, c=9

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We have formula of Discriminant:

★D=b²-4ac

D=(-2)²-4(2)(9)

D=4-72

D=-68

Here value of D comes out to be -ve.

So there is no real roots in the given quadratic equation.

Please note that here D refers to Discriminant.

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ADDITIONAL INFORMATION:

Quadratic formula:-

It is a formula to find value of roots of the quadratic equation when middle term can't be splitted into the factors.

It is expressed as:

${\underline{\boxed{\sf{X=\dfrac{-b\pm\sqrt {D}}{a}}}}}$

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Answer 2

Answer:

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