ax+b^2=a^2 bx ⇒ ax+b^2 a^2+bx=0 ⇒ ax+bx (a^2 b^2)=0 ⇒ (a+b)x (a b)(a+b)=0 ⇒ (a+b)(x (a b))=0 ⇒ (a+b)≠ 0, x=a b ∴ x=a b ...

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Roots of Quadratic Equation

Solve: ax+b2...

Question

Solve: $a x + b^{2} = a^{2} - b x$

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Solution

$a x + b^{2} = a^{2} - b x$
$\Rightarrow a x + b^{2} - a^{2} + b x = 0$
$\Rightarrow a x + b x - (a^{2} - b^{2}) = 0$
$\Rightarrow (a + b) x - (a - b) (a + b) = 0$
$\Rightarrow (a + b) (x - (a - b)) = 0$
$\Rightarrow (a + b) \neq 0, x = a - b$
$∴ x = a - b$

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a (x + y) + b (x - y) = a^{2} - a b + b^{2}

a (x + y) - b (x - y) = a^{2} + a b + b^{2}

a (x + y) + b (x - y) = a^{2} - a b + b^{2}

a (x + y) - b (x - y) = a^{2} + a b + b^{2}

solve for x and y

a(x+y)+b(x-y)=a²-ab+b²

a(x+y)-b(x-y)=a²+ab+b²

x

y

a x + b y = 1

b x + a y = \frac{(a + b)^{2}}{a^{2} + b^{2}} - 1

b x + a y = \frac{2 a b}{a^{2} + b^{2}}

\frac{a}{b x} - \frac{b}{a x} = a^{2} - b^{2}

x

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