Answer : d < 0 then it has no real roots Given : 4x² + 5x + 2 = 0 To find : Nature of the roots of the equation Solution : 》4x² + 5x + 2 = 0 As we know that D = b² – 4ac Where , a is 4 b is 5 c is 2 》D = b² – 4ac 》D = (5)² – 4(4)(2) 》D = 25 – 32 》D = – 7 Verification : D = b² – 4ac where , a is 4 , b is 5 , c is 2 and d is 7 》 D = b² – 4ac 》-7 = (5)² – 4(4)(2) 》-7 = 25 – 32 》-7 = -7 Hence verified Hence, d < 0 then it has no real roots and unique Reply
Given : [tex]\sf{4x}^{2}\:{+\:5x\:+\:2\:=\:0}[/tex] Find : Nature of the roots of the equation. [tex]{ }[/tex] [tex]\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━━━━━━[/tex] [tex]{ }[/tex] We Know that : [tex]\sf{\red{D\:=\:b}^{\red 2}\:{\red -\:\red 4 \red a \red c}}[/tex] [tex]{ }[/tex] Where, a = 4 b = 5 c = 2 [tex]{ }[/tex] Now, putting the value in formula, [tex]\sf{\pink{D\:=\:b}^{\pink 2}\:{\pink -\:\pink 4 \pink a \pink c}}[/tex] [tex]{ }[/tex] [tex]\:\:\:\:\:\:{\dashrightarrow\:\sf{D\:=\:5}^{2}\:{-\:4\:(4)\:(2)}}[/tex] [tex]{ }[/tex] [tex]\:\:\:\:\:\:{\dashrightarrow\:\sf{D\:=\:25\:-\:32}}[/tex] [tex]{ }[/tex] [tex]\:\:\:\:\:\:{\dashrightarrow\:\sf{D\:=\:-7}}[/tex] [tex]{ }[/tex] [tex]\:\therefore\:{\underline{\sf{Hence,\:d\:<\:0\:then\:it\:has\:{\textsf{\textbf{no\:real\:roots\:and\:unique. }}}}}}.[/tex] [tex]{ }[/tex] [tex]\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━━━━━━[/tex] [tex]{ }[/tex] [tex]\:\:{\underline{\bold{\purple{\sf{V\:E\:R\:I\:F\:I\:C\:A\:T\:I\:O\:N\::}}}}}[/tex] [tex]{ }[/tex] [tex]\sf{\pink{D\:=\:b}^{\pink 2}\:{\pink -\:\pink 4 \pink a \pink c}}[/tex] [tex]{ }[/tex] ★ So, know we know the value of D is –7. [tex]{ }[/tex] [tex]\:\:\:\:\:\:\leadsto\:{\sf{-7\:=\:(5)}^{2}\:{-\:4\:(4)\:(2)}}[/tex] [tex]{ }[/tex] [tex]\:\:\:\:\:\:\leadsto\:{\sf{-7\:=\:25\:-\:32}}[/tex] [tex]{ }[/tex] [tex]\:\:\:\:\:\:\leadsto\:{\sf{-7\:=\:-7}}[/tex] [tex]{ }[/tex] [tex]\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{\underline{\textsf{\textbf{Hence\:Verified!!}}}}[/tex] [tex]{ }[/tex] [tex]\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━━━━━━[/tex] [tex]{ }[/tex] Reply
Answer :
Given :
To find :
Solution :
》4x² + 5x + 2 = 0
As we know that
Where ,
》D = b² – 4ac
》D = (5)² – 4(4)(2)
》D = 25 – 32
》D = – 7
Verification :
where , a is 4 , b is 5 , c is 2 and d is 7
》 D = b² – 4ac
》-7 = (5)² – 4(4)(2)
》-7 = 25 – 32
》-7 = -7
Hence verified
Hence, d < 0 then it has no real roots and unique
Given :
Find :
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[tex]\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━━━━━━[/tex]
[tex]{ }[/tex]
We Know that :
[tex]{ }[/tex]
Where,
[tex]{ }[/tex]
Now, putting the value in formula,
[tex]{ }[/tex]
[tex]\:\:\:\:\:\:{\dashrightarrow\:\sf{D\:=\:5}^{2}\:{-\:4\:(4)\:(2)}}[/tex]
[tex]{ }[/tex]
[tex]\:\:\:\:\:\:{\dashrightarrow\:\sf{D\:=\:25\:-\:32}}[/tex]
[tex]{ }[/tex]
[tex]\:\:\:\:\:\:{\dashrightarrow\:\sf{D\:=\:-7}}[/tex]
[tex]{ }[/tex]
[tex]\:\therefore\:{\underline{\sf{Hence,\:d\:<\:0\:then\:it\:has\:{\textsf{\textbf{no\:real\:roots\:and\:unique. }}}}}}.[/tex]
[tex]{ }[/tex]
[tex]\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━━━━━━[/tex]
[tex]{ }[/tex]
[tex]\:\:{\underline{\bold{\purple{\sf{V\:E\:R\:I\:F\:I\:C\:A\:T\:I\:O\:N\::}}}}}[/tex]
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★ So, know we know the value of D is –7.
[tex]{ }[/tex]
[tex]\:\:\:\:\:\:\leadsto\:{\sf{-7\:=\:(5)}^{2}\:{-\:4\:(4)\:(2)}}[/tex]
[tex]{ }[/tex]
[tex]\:\:\:\:\:\:\leadsto\:{\sf{-7\:=\:25\:-\:32}}[/tex]
[tex]{ }[/tex]
[tex]\:\:\:\:\:\:\leadsto\:{\sf{-7\:=\:-7}}[/tex]
[tex]{ }[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{\underline{\textsf{\textbf{Hence\:Verified!!}}}}[/tex]
[tex]{ }[/tex]
[tex]\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━━━━━━[/tex]
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