Question

$\bold{x+y=\frac{7}{2}}$
$\bold{xy=\frac{5}{2}}$
$\bold{\blue{Then\:find\:the\:value\:of\:}x²-y²\:\blue{and}\: x-y}$​

Answer 1

$\huge\bold{\color{aqua}{Solution :-}}$

$\bold{\purple{\text{ it's given in the question }}}$

$\qquad\bold{\color{blue}{x+y=\color{yellow}{\frac{7}{2}}\quad \color{green}{;} \quad\color{blue}{ xy=}\color{yellow}{\frac{5}{2}}}}$

$\bold{\purple{\text{To find the value of }}}$

$\qquad\bold{\color{red}{x-y }\quad \color{green}{\& }\color{red}{\quad x^2-y^2}}$

$\purple{\bold{\text{We know that}}}$

$\quad\orange{\bold{(x-y)^2=(x+y)^2-4xy}}$

$\purple{\bold{\text{On substituting values}}}$

$\Rightarrow\bold{(x-y)^2=\big(\frac{7}{2}\big)^2-4\big(\frac{5}{2}\big)}$

$\Rightarrow\bold{(x-y)^2=\big(\frac{7^2}{2^2}\big)-\red{\cancel{\blue{4}}}^{\green{2}}\big(\frac{5}{\red{\cancel{\blue{2}}}}\big)}$

$\Rightarrow\bold{(x-y)^2=\frac{49}{4}-2×5}$

$\Rightarrow\bold{(x-y)^2=\frac{49}{4}-10}$

$\Rightarrow\bold{(x-y)^2=\frac{49-40}{4}}$

$\Rightarrow\bold{(x-y)^2=\frac{9}{4}}$

$\Rightarrow\bold{x-y=\pm\sqrt{\frac{9}{4}}}$

$\Rightarrow\bold{\green{\boxed{\red{x-y}=\pink{\pm\frac{3}{2}}}}}$

$\purple{\bold{\text{We know that}}}$

$\quad\orange{\bold{x^2-y^2=(x+y)(x-y)}}$

$\purple{\bold{\text{On substituting values}}}$

$\Rightarrow\bold{x^2-y^2=\frac{7}{2}×\bigg(\pm\frac{3}{2}\bigg)}$

$\Rightarrow\bold{x^2-y^2=\pm\frac{3×7}{2×2}}$

$\Rightarrow\bold{\green{\boxed{\red{x^2-y^2}=\pink{\pm\frac{21}{4}}}}}$