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# Find the Quadratic Polynomial whose zeros are $5+\sqrt{2}$ and $5-\sqrt{2}$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 numberC.$k\{{{x}^{2}}-10x+23\}$ , where k is any non-zero real number D.$k\{{{x}^{2}}-14x+23\}$ , where k is any non-zero real number

Last updated date: 18th Jun 2024
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Hint: For any quadratic equation of the form $a{{x}^{2}}+bx+c=0$ , which has 2 roots $\alpha$ and $\beta$
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

Complete step-by-step answer:
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) $k\{{{x}^{2}}-10x+23\}$
So, the correct answer is “Option C”.

Note: Make sure that you replace $b=-(\alpha +\beta )a$ sometimes we forget to take that negative sign,
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 $\alpha \beta =\dfrac{c}{a}$