Question

1) a⁴ + 4a²-5
2) x²y²+ 23xy - 420
please Answer
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Answer 1

Question :

Solve :

1) a⁴ + 4a²-5

2) x²y²+ 23xy - 420

Solution :

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

Let a²= t , then

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

By Quadratic Formula

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

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

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

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

$\sf\implies\:t=\dfrac{-4\pm6}{2}$

$\sf\implies\:t=1\:or-1$

t can't be -1 , then

$\sf\implies\:a^2=1$

$\sf\implies\:a^2-1=0$

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

$\sf\implies\:a=\pm1$

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

Let x²y² = t , then

$\sf=t+23xy-420$

By Quadratic Formula

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

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

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

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

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

$\sf\implies\:t=\dfrac{-23\pm47}{2}$

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

$\sf\implies\:t=12\:or-35$

t can't be -35

$\sf\implies\:x^2y^2=12$

$\sf\implies\:x^2y^2-12=0$

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

$\sf\implies\:xy=\pm\sqrt{12}$

Answer 2

Answer:

Solution-1:-

Given Polynomial

$\sf a^4+4a^2-5$

• We have to find a

Let $a^2=x$

• Then convert it as a equation

$\quad {:}\longrightarrow\tt x^2+4x-5=0$

• Use midterm factorisation

$\quad {:}\longrightarrow\tt x^2+5x-x-5=0$

• Factorise

$\quad {:}\longrightarrow\tt x (x+5)-1 (x+5)=0$

$\quad {:}\longrightarrow\tt (x+5)(x-1)=0$

$\quad {:}\longrightarrow\tt x+5=0\quad or\quad x-1=0$

$\quad {:}\longrightarrow\tt x=-5\quad or \quad x=1$

• a can't be -5
• thus

$\quad {:}\longrightarrow\tt a^2=x$

$\quad {:}\longrightarrow\tt a^2 =1$

$\quad {:}\longrightarrow\tt a^2-1=0$

• Use algebraic identity

$\boxed{\sf a^2-b^2=(a+b)(a-b)}$

$\quad {:}\longrightarrow\tt (a)^2-(1)^2=0$

$\quad {:}\longrightarrow\tt (a+1)(a-1)=0$

$\quad {:}\longrightarrow\tt a+1=0\quad or\quad a-1=0$

$\quad {:}\longrightarrow\tt a=-1 \:or\: a=1$

$\therefore\sf a=\underline{+}1$

Solution-2:-

Given Polynomial

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

Let

$\sf x^2y^2=d$

• convert it as equation

$\quad {:}\longrightarrow\tt d+23xy-420=0$

• substract -23xy from both sides

$\quad {:}\longrightarrow\tt d+\cancel {23xy}-\cancel{23xy}=0-23xy$

$\quad {:}\longrightarrow\tt d-420=-23xy$

• Add 420 on both sides

$\quad {:}\longrightarrow\tt d-420+420=-23xy+420$

• Simplify

$\quad {:}\longrightarrow\tt d=-23xy+420$