Question

What two numbers when multiplied equal 8 and when added equal 10

Answer 1

Question: Two numbers when multiplied are equal to 8, and when added are equal to 10.

Solution:

Let us consider both the numbers to be 'x' and 'y'.

ATQ,

$\Longrightarrow \sf Product \ of \ the \ two \ digits = 8\\ \\\Longrightarrow \sf x \times y = 8\\ \\\Longrightarrow \sf xy = 8 \ \ \longmapsto \textcircled{\scriptsize \sf{1}}$

Also,

$\Longrightarrow \sf Sum \ of \ both \ the \ digits = 10\\ \\\Longrightarrow \sf x + y = 10\\ \\\Longrightarrow \sf {y = 10 - x} \ \ \longmapsto \textcircled{\scriptsize \sf {2}}$

$\sf Substituting \ \textcircled{\scriptsize \sf{2}} \ in \ \textcircled{\scriptsize \sf{1}} \ we \ get,\\ \\ \\\Longrightarrow \sf \ xy = 8\\ \\\Longrightarrow \sf \ x\left(10-x\right)=8\\ \\ \Longrightarrow \sf \ 10x-x^2=8\\ \\ \Longrightarrow \sf \ 10x-x^2-8=0\\ \\ \Longrightarrow \sf \ -x^2+10x-8=0\\ \\ \\ \sf Multiplying \ by \ negative \ sign \ on \ both \ sides \ we \ get,\\ \\ \\ \Longrightarrow \sf \ x^2-10x+8=0$

Using the Quadratic formula we get,

$\boxed{\sf QUADRATIC \ FORMULA: x = \dfrac{-b \pm \sqrt{b^2-4ac \ } }{2a}}$

In the case of this equation,

b = -10

a = 1

c = 8

Therefore,

$\Longrightarrow \ \sf x = \dfrac{-b \pm \sqrt{b^2-4ac \ } }{2a}\\ \\ \\\Longrightarrow \ \sf x = \dfrac{-(-10) \pm \sqrt{(-10)^2-4(1)(8) \ } }{2(1)}\\ \\ \\\Longrightarrow \ \sf x = \dfrac{10 \pm \sqrt{100-32 \ } }{2}\\ \\ \\\Longrightarrow \ \sf x = \dfrac{10 \pm \sqrt{68 \ } }{2}\\ \\ \\\Longrightarrow \ \sf x = \dfrac{10 \pm \sqrt{68 \ } }{2}\\ \\ \\\Longrightarrow \ \sf x = \dfrac{10 + \sqrt{68 \ } }{2} \ , \ \dfrac{10 - \sqrt{68 \ } }{2}$

$\Longrightarrow \ \sf x = \dfrac{10 + 2\sqrt{17 \ } }{2} \ , \ \dfrac{10 - 2\sqrt{17 \ }}{2}\\ \\ \\\Longrightarrow \ \sf x = \dfrac{2 \! \! \!{/} \left(5+\sqrt{17}\right)}{2 \! \! \!{/}} \ , \ \dfrac{2 \! \! \!{/} \left(5 - \sqrt{17}\right)}{2\! \! \!{/}}\\ \\ \\\Longrightarrow \ \sf x = 5+\sqrt{17} \ , \ 5 - \sqrt{17}\right$

Since we have two values for x, we'll have two cases, Case I and Case II.

$\sf \underline{Case \ I}\\ \\When \ x = 5 + \sqrt{17}\\ \\\sf Substitute \ the \ value \ of \ 'x' \ in \ Equation \ \textcircled{\scriptsize \sf {2}}\\ \\\sf \Longrightarrow y = 10 - x \\ \\\sf \Longrightarrow y = 10 - (5+\sqrt{17}) \\ \\\sf \Longrightarrow y = 10 - 5-\sqrt{17} \\ \\\sf \Longrightarrow y = 5-\sqrt{17} \\ \\ \\\sf Now,\\ \\\sf \Longrightarrow x + y = 10\\ \\\sf \Longrightarrow 5 + \bcancel{\sqrt{17}} + 5 \bcancel{- \sqrt{17}} = 10\\ \\\sf \Longrightarrow 10 = 10\\ \\ \\Hence \ Solved.$

$\sf \underline{Case \ II}\\ \\When \ x = 5 - \sqrt{17}\\ \\\sf Substitute \ the \ value \ of \ 'x' \ in \ Equation \ \textcircled{\scriptsize \sf {2}}\\ \\\sf \Longrightarrow y = 10 - x \\ \\\sf \Longrightarrow y = 10 - (5-\sqrt{17}) \\ \\\sf \Longrightarrow y = 10 - 5+\sqrt{17} \\ \\\sf \Longrightarrow y = 5+\sqrt{17}\\ \\ \\\sf Now,\\ \\\sf \Longrightarrow x + y = 10\\ \\\sf \Longrightarrow 5 \bcancel{- \sqrt{17}} + 5 + \bcancel{\sqrt{17}} = 10\\ \\\sf \Longrightarrow 10 = 10\\ \\ \\Hence \ Solved.$

Therefore both cases are possible.

$\sf \underline{\underline{SOLUTION}}:\\ \\\Longrightarrow \sf If \ x = 5 + \sqrt{17} \ , \ y = 5 - \sqrt{17}\\ \\ \\\Longrightarrow \sf If \ x = 5 - \sqrt{17} \ , \ y = 5 + \sqrt{17}$

Answer 2

$\tt\sc Let \ the\ numbers \ be \ x \ and \ y\ .\\\\\tt Product\ of\ two \ numbers=8\\\\\tt Sum \ of \ two \ numbers = 10\\\\x+y=10\\\\\longrightarrow y=10-x\\\\\tt xy=8\\\\\tt\longrightarrow x(10-x)=8\\\\\tt\longrightarrow 10x-x^{2} =8\\\\\tt\longrightarrow 10x-x^{2} -8=0\\\\\tt\longrightarrow x^{2} -10x+8=0\\\\\tt\bf x= \dfrac{-b \pm \sqrt{b^2-4ac \ } }{2a}\\\\\\\tt\longrightarrow x = \dfrac{-(-10) \pm \sqrt{(-10)^2-4\times1\times8 \ } }{2\times1}$

$\tt\longrightarrow x = \dfrac{10 \pm \sqrt{100-32 \ } }{2}$

$\tt\longrightarrow x = \dfrac{10 \pm \sqrt{68 \ } }{2}$

$\tt\longrightarrow x = \dfrac{10\pm\sqrt{17\times4} }{2}$

$\tt\longrightarrow x=\dfrac{10\pm2\sqrt{17} }{2}$

$\tt\longrightarrow x=\dfrac{10+2\sqrt{17} }{2} \ ,\ x=\dfrac{10-2\sqrt{17} }{2}$

$\tt\longrightarrow x=\dfrac{2\times(5+\sqrt{17}) }{2} \ ,\ x= \dfrac{2\times(5-\sqrt{17} )}{2}$

$\tt\longrightarrow x=5+\sqrt{17} \ , \ x=5-\sqrt{17}$

$\tt If \ x = 5+\sqrt{17} ,y = 10-(5+\sqrt{17} )=\bf 5-\sqrt{17} \\\\\tt If \ x = 5-\sqrt{17} \ , \ y=10-(5-\sqrt{17}) =\bf 5+\sqrt{17}$

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