Question

+ 4a²-5
2) x²y²+ 23xy - 420
please Answer
just solution ​

Answer 1

Answer:

y = 2, 2.5. x² – 39x + 324 = 0 y² – 47y + 420 = 0. A. X > Y B. X < Y C. X ≥ Y D. X ≤ Y E. X = Y or relation cannot ...

Answer 2

Question :

Solve :

1) a⁴ + 4a²-5

2) x²y²+ 23xy - 420

Solution :

1) $\sf\:a^4+4a^2-5a$

Let a²= t , then

$\sf=t^2+4t-5=t$

By Quadratic Formula

$\sf\:x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}x=$

$\sf\implies\:t=\dfrac{-4\pm\sqrt{4^2-4(-5)}}{2}⟹t=$

$\sf\implies\:t=\dfrac{-4\pm\sqrt{16+20}}{2}⟹t=$

$\sf\implies\:t=\dfrac{-4\pm\sqrt{36}}{2}⟹t=$

$\sf\implies\:t=\dfrac{-4\pm6}{2}⟹t=$

$\sf\implies\:t=1\:or-1⟹t=1or−1$

t can't be -1 , then

$\sf\implies\:a^2=1⟹a$

$\sf\implies\:a^2-1=0⟹a$

$\sf\implies\:(a+1)(a-1)=0⟹(a+1)(a−1)=0$

$\sf\implies\:a=\pm1⟹a=±1$

2) $\sf\:x^2y^2+23xy-420x$

Let x²y² = t , then

\sf=t+23xy-420=t+23xy−420

By Quadratic Formula

$\sf\:x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}x=$

$\sf\implies\:t=\dfrac{-23\pm\sqrt{23^2-4(-420)}}{2}⟹t=$

$\sf\implies\:t=\dfrac{-23\pm\sqrt{529+1680}}{]}⟹t=$

$\sf\implies\:t=\dfrac{-23\pm\sqrt{2209}}{2}⟹t=$

$\sf\implies\:t=\dfrac{-23\pm\sqrt{529+1680}}{2}⟹t=$

[/tex]\sf\implies\:t=\dfrac{-23\pm47}{2}⟹t=[/tex]

$\sf\implies\:t=\dfrac{-23+47}{2}\:or\dfrac{-23-47}{2}⟹t=$

$\sf\implies\:t=$

$12\:or-35⟹t=$

t can't be -35

$\sf\implies\:x^2y^2=12⟹x$

$\sf\implies\:x^2y^2-12=0⟹x$

$\sf\implies\:(xy+\sqrt{12})(xy-\sqrt{12})=0⟹(xy+$

$\sf\implies\:xy=\pm\sqrt{12}⟹xy=±$