Question

find the nature of roots of the equation 2 x square - 3 x + 5 is equal to zero ​

Answer 1

Answer:

Step-by-step explanation:

Hey there!

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2 x^2 - 3x + 5 = 0

Where,

a = 2

b = -3

c = 5

[ According to ax^2 + bx + c = 0 , we just compared the equation with this]

Now, we know,

D = b^2 - 4ac

D = (-3)^2 - 4 × 2 × 5

D = 9 - 8 × 5

D = 9 - 40

D = - 31

Now, we know,

If D is less than 0 ( D < 0 ) then no real roots exist.

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I Hope it's right and hope it helps you...!!!

Answer 2

$\huge\mathfrak{solution}$

$2 {x}^{2} - 3x + 5 = 0 \\ \\ Now ,\\ \: comparing \:the \:quadratic \\ equation \: 2 {x}^{2} - 3x + 5 \: with \: the \: \\ quadratic \: equation \\ a {x}^{2} + bx + c = 0 \: , we \: get$

$a = 2 \: \: \: b = - 3 \: \: \: c= 5$

$The \: discriminant = {b}^{2} - 4ac \\ = ( { - 3)}^{2} - 4.2.5 \\ = 9 - 40 \\ = - 31 \\ We \: will \: not \: get \: any \: real \: from \: \\ the \: given \: quadratic \: equation$

$\mathfrak{hope \: it \: helps \: uhh}$