2*4=8,3*5=11,7*9=23,9*7=25, 5*4=?
Answers
Answer:
5 * 4 = 14
Step-by-step explanation:
Given that;
2 * 4 = 8
3 * 5 = 11
7 * 9 = 23
9 * 7 = 25
5 * 4 = ?
Let's see the sequence of the given data. We will solve this question by following the steps :
1. Add the two numbers first and then add the resultant number with the first number. Example :
( 2 + 4 ) = 6
( 6 + 2 ) = 8
Hence, 8 is the required number.
2. Again, do the same with the rest of the numbers.
( 3 + 5 ) = 8
( 8 + 3 ) = 11
Hence, 11 is the required number.
3. Follow the same again ;
( 7 + 9 ) = 16
( 16 + 7 ) = 23
Hence, 23 is the required number.
4. The last numbers are ;
( 9 + 7 ) = 16
( 16 + 9 ) = 25
Hence, 25 is the required number.
5. Now, the required answer for last term is in the same form ;
( 5 + 4 ) = 9
( 9 + 5 ) = 14
Hence, 14 is the required number.
Step-by-step explanation:
\huge{\underline{\bold{Answer:}}}
Answer:
\sf x = \dfrac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} - \sqrt{p - 2q} } x=
p+2q
−
p−2q
p+2q
+
p−2q
So, let's rationalise RHS first,
\sf \frac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} - \sqrt{p - 2q} } \times \frac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} + \sqrt{p - 2q} }
p+2q
−
p−2q
p+2q
+
p−2q
×
p+2q
+
p−2q
p+2q
+
p−2q
\sf \dfrac{ (\sqrt{p + 2q} + \sqrt{p - 2q} ) {}^{2} }{ (\sqrt{p + 2q} ) {}^{2} - ( \sqrt{p - 2q}) {}^{2} }
(
p+2q
)
2
−(
p−2q
)
2
(
p+2q
+
p−2q
)
2
\sf \frac{ (\sqrt{p + 2q}) {}^{2} + ( \sqrt{p - 2q}) {}^{2} + 2(\sqrt{p + 2q})(\sqrt{p - 2q})}{ \cancel p + 2q - \cancel p + 2q}
p
+2q−
p
+2q
(
p+2q
)
2
+(
p−2q
)
2
+2(
p+2q
)(
p−2q
)
\sf \frac{p + 2q + p - 2q + 2 \sqrt{p + 2q} \times \sqrt{p - 2q} }{4q}
4q
p+2q+p−2q+2
p+2q
×
p−2q
\sf \frac{2p + 2 \sqrt{p + 2q} \times \sqrt{p - 2q}}{4q}
4q
2p+2
p+2q
×
p−2q
\begin{gathered}\sf x = \frac{2(p + \sqrt{p + 2q} \times \sqrt{p - 2q})}{2(2q)} \\ \\ \sf x = \frac{p +\sqrt{p + 2q} \times \sqrt{p - 2q}}{2q} \\ \\ \bf on \: cross \: multiplying : - \\ \\ \sf 2qx = p + \sqrt{p + 2q} \times \sqrt{p - 2q} \\ \\ \sf 2qx - p = \sqrt{p + 2q} \times \sqrt{p - 2q} \\ \\ \bf on \: squaring \: both \: sides : \\ \\ \sf (2qx - p) {}^{2} = (\sqrt{p + 2q} \times \sqrt{p - 2q}) {}^{2} \end{gathered}
x=
2(2q)
2(p+
p+2q
×
p−2q
)
x=
2q
p+
p+2q
×
p−2q
oncrossmultiplying:−
2qx=p+
p+2q
×
p−2q
2qx−p=
p+2q
×
p−2q
onsquaringbothsides:
(2qx−p)
2
=(
p+2q
×
p−2q
)
2
\begin{gathered} \sf{4q {}^{2} x {}^{2} + p {}^{2} - 4qxp= (p + 2q)(p - 2q)} \\ \\ \sf 4q {}^{2} x {}^{2} + p {}^{2} - 4qxp =p {}^{2} - 4q {}^{2} \\ \\ \sf 4q {}^{2} x {}^{2} + p {}^{2} - 4qxp - p {}^{2} + 4q {}^{2} = 0 \\ \\ \sf 4q {}^{2} x{}^{2} - 4qxp + 4q {}^{2} = 0 \\ \\ \sf 4q(qx {}^{2} - xp + q) = 0 \\ \\ \boxed{\bf qx {}^{2} - xp + q= 0}\end{gathered}
4q
2
x
2
+p
2
−4qxp=(p+2q)(p−2q)
4q
2
x
2
+p
2
−4qxp=p
2
−4q
2
4q
2
x
2
+p
2
−4qxp−p
2
+4q
2
=0
4q
2
x
2
−4qxp+4q
2
=0
4q(qx
2
−xp+q)=0
qx
2
−xp+q=0
\huge{\underline{\bold{PROVED.}}}
PROVED.