Question

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sʜᴏᴡ ᴛʜᴀᴛ ᴛʜᴇ ғᴏʟʟᴏᴡɪɴɢ ᴇϙᴜᴀᴛɪᴏɴ ʜᴀs ʀᴇᴀʟ ʀᴏᴏᴛs, ᴀɴᴅ sᴏʟᴠᴇ ᴇᴀᴄʜ ʙʏ ᴜsɪɴɢ ᴛʜᴇ ϙᴜᴅʀᴀᴛɪᴄ ғᴏʀᴍᴜʟᴀ. 

${x}^{2} - 4x - 1 = 0$
Ćĺâśš 10 
զմαժɾαԵíc ҽզմαԵíօղs


ͲᎻᎪΝᏦᎽϴႮ

Answer 1

Hey mate,

We know,

For real roots,

D>= 0

So,

D = b^2 - 4ac

= 16 + 4

= 20

So,

20 is >=0

So the equation has real root.

For roots,

X = - b +- root(D) /2a

= 4 +-2root(5)/2

= 2+-1root(5)

So roots are 2 + 1 root(5) and 2 - 1 root(5)

Hope this helps you out!

Answer 2

For any equation to have real root the discriminant should be greater than or equal to 0 .

In the given equation :-

${x}^{2} - 4x - 1 = 0$

a=1 , b= -4 and c=-1

$\textbf{Discriminant(D) = {b}^{2} -4ac }$

$= ( - {4}^{2} ) - 4 \times 1 \times - 1$

=16 + 4

= 20

Since, D = 20 is greater than 0 , so it has two unequal real roots.

${ \boxed{ \bf{quadratic \: formula = \frac{b + \sqrt{d} }{2a} and \: \frac{b - \sqrt{d} }{2a} }}}$

So, the two roots are :-

$\bf{ \frac{ -(- 4 )+ \sqrt{20} }{2 \times 1} and \: \frac{ -(- 4 )- \sqrt{20} }{2 \times 1} }$

$\bf{ \frac{ 4 + 2 \sqrt{5} }{2} \: an d \: \frac{ 4 - 2 \sqrt{5} }{2} }$

$= \bf{ \frac{2( 2 + \sqrt{5} )}{2} \: and \: \frac{2( 2 - \sqrt{5}) }{2} }$

$= \bf{ \frac{ \cancel2( \sqrt{5 } +2) }{ \cancel2} \: and \: } \frac{ \cancel2( 2 - \sqrt{5}) }{ \cancel2}$

$\underline{\textbf{So, the two roots are}}$:-

$\boxed{ \bf{ \sqrt{5} + 2 \: and \: 2 - \sqrt{5} }}$

$\huge {\bf{hope \: it \: helps : )}}$