/
In Copying a second degree equation
from the blackboard a student made
a mistake and wrote the Coefficient
of x
as 7 instead of -7. He found
the solutions are - 4 and - 3
a )what was the correct equations
b) what are the solutions
Answers
Answer:
Answer:
. \huge \underline \mathscr \blue{answer}.
answer
⇢(Diagonal)
2
=(Length)
2
+(Breadth)
2
⇢(BD)
2
=(BC)
2
+(CD)
2
⇢(BD)
2
=(24cm)
2
+(7cm)
2
⇢(BD)
2
=576cm
2
+49cm
2
⇢(BD)
2
=625cm
2
⇢BD=
625cm
2
⇢BD=
25cm×25cm
⇢
BD=25cm
⎩
⎪
⎪
⎪
⎧
Diagonal
⎭
⎪
⎪
⎪
⎫
⇢(Diagonal)2=(Length)2+(Breadth)2⇢(BD)2=(BC)2+(CD)2⇢(BD)2=(24cm)2+(7cm)2⇢(BD)2=576cm2+49cm2⇢(BD)2=625cm2⇢BD=625cm2⇢BD=25cm×25cm⇢BD=25cm⎩⎪⎪⎪⎧Diagonal⎭⎪⎪⎪⎫⠀∴
Hence, Length of Diagonal is C) 25 cm
.∴Hence,LengthofDiagonalisC)25cm.
\begin{gathered}\begin{gathered}\dashrightarrow\sf\:\:(Diagonal)^2=(Length)^2+(Breadth)^2\\\\\\\dashrightarrow\sf\:\:(BD)^2=(BC)^2+(CD)^2\\\\\\\dashrightarrow\sf\:\:(BD)^2=(24\:cm)^2+(7\:cm)^2\\\\\\\dashrightarrow\sf\:\:(BD)^2=576\:cm^2+49\:cm^2\\\\\\\dashrightarrow\sf\:\:(BD)^2=625\:cm^2\\\\\\\dashrightarrow\sf\:\:BD=\sqrt{625\:cm^2}\\\\\\\dashrightarrow\sf\:\:BD=\sqrt{25\:cm \times 25\:cm}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf BD=25\:cm}}\qquad\bigg\lgroup\bf Diagonal\bigg\rgroup\end{gathered} ⇢(Diagonal) 2 =(Length) 2 +(Breadth) 2 ⇢(BD) 2 =(BC) 2 +(CD) 2 ⇢(BD) 2 =(24cm) 2 +(7cm) 2 ⇢(BD) 2 =576cm 2 +49cm 2 ⇢(BD) 2 =625cm 2 ⇢BD= 625cm 2 ⇢BD= 25cm×25cm ⇢ BD=25cm ⎩ ⎪ ⎪ ⎪ ⎧ Diagonal ⎭ ⎪ ⎪ ⎪ ⎫ ⠀ \therefore\:\underline{\textsf{Hence, Length of Diagonal is C) \textbf{25 cm}}}.∴ Hence, Length of Diagonal is C) 25 cm .\end{gathered}
⇢(Diagonal)
2
=(Length)
2
+(Breadth)
2
⇢(BD)
2
=(BC)
2
+(CD)
2
⇢(BD)
2
=(24cm)
2
+(7cm)
2
⇢(BD)
2
=576cm
2
+49cm
2
⇢(BD)
2
=625cm
2
⇢BD=
625cm
2
⇢BD=
25cm×25cm
⇢
BD=25cm
⎩
⎪
⎪
⎪
⎧
Diagonal
⎭
⎪
⎪
⎪
⎫
⇢(Diagonal)2=(Length)2+(Breadth)2⇢(BD)2=(BC)2+(CD)2⇢(BD)2=(24cm)2+(7cm)2⇢(BD)2=576cm2+49cm2⇢(BD)2=625cm2⇢BD=625cm2⇢BD=25cm×25cm⇢BD=25cm⎩⎪⎪⎪⎧Diagonal⎭⎪⎪⎪⎫⠀∴
Hence, Length of Diagonal is C) 25 cm
.∴Hence,LengthofDiagonalisC)25cm.