Question

The product of LCM and HCF of two numbers is 48 and their difference is 8. Then the numbers are

Answer 1

SOLUTION :

$\underline{\bf{Given\::}}}}$

The product of L.C.M and H.C.F of two number is 48 & their difference is 8.

$\underline{\bf{Explanation\::}}}}$

Let the two number be r & m

$\underbrace{\sf{According\:to\:the\:question\::}}}}$

$\longrightarrow\sf{r-m=8}\\\\\longrightarrow\sf{r=8+m....................(1)}$

&

We know that;

$\boxed{\bf{Product\:of\:two\:numbers=H.C.F\times L.C.M}}}}}$

$\longrightarrow\sf{rm=48}\\\\\longrightarrow\sf{(8+m)m=48\:\:\:[from(1)]}\\\\\longrightarrow\sf{8m+m^{2} =48}\\\\\longrightarrow\sf{m^{2} +8m-48=0}$

By using quadratic Formula :

$\boxed{\bf{\frac{-b\pm\sqrt{b^{2}-4ac } }{2a} }}}}$

As we know that given equation compared with ax² + bx + c;

• a = 1
• b = 8
• c = -48

Now;

$\longrightarrow\sf{x=\dfrac{-8\pm\sqrt{(8)^{2} -4\times 1\times (-48)} }{2\times 1} }\\\\\\\longrightarrow\sf{x=\dfrac{-8\pm\sqrt{64+192} }{2}} \\\\\\\longrightarrow\sf{x=\dfrac{-8\pm\sqrt{256} }{2} }\\\\\\\longrightarrow\sf{x=\dfrac{-8\pm16}{2} }\\\\\\\longrightarrow\sf{x=\dfrac{-8+16}{2} \:\:Or\:\:x=\dfrac{-8-16}{2} }\\\\\\\longrightarrow\sf{x=\cancel{\dfrac{8}{2}} \:\:Or\:\: x=\cancel{\dfrac{-24}{2}}}\\\\\\\longrightarrow\bf{x=4\:\:Or\:\:x\neq -12}$

Putting the value of x = 4 (m) in equation (1),we get;

$\longrightarrow\sf{r=8+4}\\\\\longrightarrow\bf{r=12}$

Thus;

The numbers will be 12 & 4 .