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A. \[k\{{{x}^{2}}-2x+23\}\] , where k is any non-zero real number

B.\[k\{{{x}^{2}}-16x+23\}\] , where k is any non-zero real number

C.\[k\{{{x}^{2}}-10x+23\}\] , where k is any non-zero real number

D.\[k\{{{x}^{2}}-14x+23\}\] , where k is any non-zero real number

Answer
Verified

two things are known to us

Sum of roots means \[\alpha +\beta \] is equals to \[\dfrac{-b}{a}\] and product of roots means \[\alpha \beta \] equals to \[\dfrac{c}{a}\]

Here, a is any non-zero real number because it is in denominator so can’t be 0

We are given values of \[\alpha \] and \[\beta \] in the question so we can find desired quadratic equation

We have a general quadratic equation as \[a{{x}^{2}}+bx+c=0\] which has roots \[\alpha \] and \[\beta \]

But in the question, we have values of \[\alpha =5+\sqrt{2}\] and \[\beta =5-\sqrt{2}\]

So using formula , Sum of roots means \[\alpha +\beta \] is equals to \[\dfrac{-b}{a}\] ,\[\alpha +\beta =\dfrac{-b}{a}\]

And product of roots means \[\alpha \beta \] equals to \[\dfrac{c}{a}\], \[\alpha \beta =\dfrac{c}{a}\]

\[\alpha +\beta =5+\sqrt{2}+5-\sqrt{2}\]

\[\alpha +\beta =10....(1)\]

And \[\alpha \beta =(5+\sqrt{2})(5-\sqrt{2})\]

Now using formula \[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]

\[\alpha \beta =(5+\sqrt{2})(5-\sqrt{2})\]

After applying this formula our equation looks like

\[\alpha \beta =({{5}^{2}}-{{(\sqrt{2})}^{2}})=25-2\]

Finally, we get

\[\alpha \beta =23....(2)\]

Now we know that \[\alpha +\beta =\dfrac{-b}{a}\] and \[\alpha \beta =\dfrac{c}{a}\]

Putting values from equation (1 and 2)

We get \[10=\dfrac{-b}{a}\] and \[23=\dfrac{c}{a}\]

On solving we get \[b=-10a\] and \[c=23a\]

Now we can put this value of b and c in original equation \[a{{x}^{2}}+bx+c=0\]

We get \[a{{x}^{2}}-10ax+23a=0\] , now in this equation we can take a common, then it will look like

\[a({{x}^{2}}-10x+23)=0\], don’t confuse with k

A or k is just a random number which is non zero real number

Hence answer is (C) \[k\{{{x}^{2}}-10x+23\}\]

And take \[b=(\alpha +\beta )a\] which will results in wrong answer , if in a question we are given equation and asking for solution then also we can apply the same procedure as mentioned above and use theses formulas \[\alpha +\beta =\dfrac{-b}{a}\] and \[\alpha \beta =\dfrac{c}{a}\]