If a= 2, b=3 find the values of (a to the power a + b to the power b) whole into -1
Answers
Step-by-step explanation:
en equation is
a^2b^2x^2-(4b^4-3a^4)x-12a^2b^2=0a
2
b
2
x
2
−(4b
4
−3a
4
)x−12a
2
b
2
=0
\text{Then, the quadratic formula is}Then, the quadratic formula is
\bf\,x=\displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{2a}x=
2a
−b±
b
2
−4ac
x=\displaystyle\frac{(4b^4-3a^4)\pm\sqrt{(4b^4-3a^4)^2+4(a^2b^2)(12a^2b^2)}}{2a^2b^2}x=
2a
2
b
2
(4b
4
−3a
4
)±
(4b
4
−3a
4
)
2
+4(a
2
b
2
)(12a
2
b
2
)
x=\displaystyle\frac{(4b^4-3a^4)\pm\sqrt{(4b^4-3a^4)^2+48a^4b^4}}{2a^2b^2}x=
2a
2
b
2
(4b
4
−3a
4
)±
(4b
4
−3a
4
)
2
+48a
4
b
4
x=\displaystyle\frac{(4b^4-3a^4)\pm\sqrt{16b^8+9a^8-24a^2b^2+48a^4b^4}}{2a^2b^2}x=
2a
2
b
2
(4b
4
−3a
4
)±
16b
8
+9a
8
−24a
2
b
2
+48a
4
b
4
x=\displaystyle\frac{(4b^4-3a^4)\pm\sqrt{16b^8+9a^8+24a^2b^2}}{2a^2b^2}x=
2a
2
b
2
(4b
4
−3a
4
)±
16b
8
+9a
8
+24a
2
b
2
x=\displaystyle\frac{(4b^4-3a^4)\pm\sqrt{(4b^4+3a^4)^2}}{2a^2b^2}x=
2a
2
b
2
(4b
4
−3a
4
)±
(4b
4
+3a
4
)
2
x=\displaystyle\frac{(4b^4-3a^4)\pm(4b^4+3a^4)}{2a^2b^2}x=
2a
2
b
2
(4b
4
−3a
4
)±(4b
4
+3a
4
)
x=\displaystyle\frac{(4b^4-3a^4)+(4b^4+3a^4)}{2a^2b^2},\;\displaystyle\frac{(4b^4-3a^4)-(4b^4+3a^4)}{2a^2b^2}x=
2a
2
b
2
(4b
4
−3a
4
)+(4b
4
+3a
4
)
,
2a
2
b
2
(4b
4
−3a
4
)−(4b
4
+3a
4
)
x=\displaystyle\frac{8b^4}{2a^2b^2},\;\displaystyle\frac{-6a^4}{2a^2b^2}x=
2a
2
b
2
8b
4
,
2a
2
b
2
−6a
4
\implies\bf\,x=\displaystyle\frac{4b^2}{a^2},\;\displaystyle\frac{-3a^2}{b^2}⟹x=
a
2
4b
2
,
b
2
−3a
2
\therefore\textbf{The solution set is \{$\frac{4b^2}{a^2},\;\frac{-3a^2}{b^2}$\}}∴The solution set is {
a
2
4b
2
,
b
2
−3a
2
}
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