Question

If $x+y=\frac{7}{2}$ and $xy=\frac{5}{2}$; find :$x^2-y^2$.

• A. $\pm \frac{11}{4}$
• B. $\pm \frac{27}{4}$
• C. $\pm \frac{13}{4}$
• D. $\pm \frac{21}{4}$

Answer 1

The correct option is D $\pm \frac{21}{4}$

GIven, $x+y=\frac{7}{2}$ and $xy=\frac{5}{2}$

$x+y=\frac{7}{2}$

Squaring both sides, we get

$=>(x+y{)}^2=(\frac{7}{2}{)}^2$

$=>x^2+y^2+2xy=\frac{49}{4}$

$=>x^2+y^2+2\times \frac{5}{2}=\frac{49}{4}$

$=>x^2+y^2+5=\frac{49}{4}$

$=>x^2+y^2=\frac{49}{4}-5$

$=>x^2+y^2=\frac{49-20}{4}$

$=>x^2+y^2=\frac{29}{4}$

Now,

$(x-y{)}^2=x^2+y^2-2xy$

$=\frac{29}{4}-2\frac{5}{2}$

$=\frac{29-20}{4}$

$=\frac{9}{4}$

$=>(x-y)=\sqrt{\frac{9}{4}}$

$=>(x-y)=\pm \frac{3}{2}$

Now,

$x^2-y^2=(x+y)(x-y)$

$=\left(\frac{7}{2}\right)(\pm \frac{3}{2})$

$=\pm \frac{21}{4}$