Question

b²x²-(a²+b²)x+a²=0 by splitting middle term​

Answer 1

Step-by-step explanation:

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b²x²-(a²+b²)x+a²=0

b²x²-a²x-b²x+a²=0

b²x²-b²x-a²x+a²=0

b²x(x-1)-a²(x-1)=0

(b²x-a²)(x-1)=0

x=a²/b²

x=1

Answer 2

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{1 \ and \ \frac{a^{2}}{b^{2}} \ are \ the \ roots \ of \ the}$

$\sf{given \ quadratic \ equation.}$

$\sf\orange{Given:}$

$\sf{The \ given \ quadratic \ equation \ is}$

$\sf{\implies{b^{2}x^{2}-(a^{2}+b^{2})x+a^{2}=0}}$

$\sf\pink{To \ find:}$

$\sf{The \ roots \ of \ the \ equation. }$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{The \ given \ quadratic \ equation \ is}$

$\sf{\implies{b^{2}x^{2}-(a^{2}+b^{2})x+a^{2}=0}}$

$\sf{\implies{b^{2}x^{2}-a^{2}x-b^{2}x+a^{2}=0}}$

$\sf{\implies{b^{2}x^{2}-b^{2}x-a^{2}x+a^{2}=0}}$

$\sf{\implies{b^{2}x(x-1)-a^{2}(x-1)=0}}$

$\sf{\implies{(x-1)(b^{2}x-a^{2})=0}}$

$\sf{\implies{\therefore{(x-1)=0 \ or \ (b^{2}x-a^{2})=0}}}$

$\sf{\implies{\therefore{x=1 \ or \ x=\frac{a^{2}}{b^{2}}}}}$

$\sf\purple{\tt{\therefore{1 \ and \ \frac{a^{2}}{b^{2}} \ are \ the \ roots \ of \ the}}}$

$\sf\purple{\tt{given \ quadratic \ equation.}}$