Question

Plz answer this question​

Answer 1

given 2 zeroes = 5+2$\sqrt3$ , 5-2$\sqrt3$

therefore , the equation must've been :

[x+-(5+2$\sqrt3$)] * [x+(5-2$\sqrt3$] = 0

apply (x+a)*(x+b)) = x^2 - (a+b)x + ab formula

u will get x^2 - 4$\sqrt3$- 13

Answer 2

$\large\boxed{\fcolorbox{blue}{yellow} {Answer}}$

${x}^{2} - 10x + 13 = 0$

Given

The roots of the quadratic equation are

$(5 + 2 \sqrt{3} )$

And

$(5 - 2 \sqrt{3} )$

To find

The quadratic equation

Solution

Let the roots of the quadratic equation be alpha and beta.

$\alpha = (5 + 2 \sqrt{3} ) \: \: \\ \beta = (5 - 2 \sqrt{3} ) \\ \\ \alpha + \beta \: = \frac{ - b}{a} \\ \\ \alpha . \beta = \frac{c}{a} \\ \\ now \\ \alpha + \beta = (5 + 2 \sqrt{3} ) + (5 - 2 \sqrt{3} ) \\ = 5 + 2 \sqrt{3} + 5 - 2 \sqrt{3} \\ = 5 + 5 \\ = 10 \\ \\ \\ now \\ \alpha . \beta = (5 + 2 \sqrt{3} ) \times (5 - 2 \sqrt{3} ) \\ = 5 \times (5 - 2 \sqrt{3} ) + 2 \sqrt{3} \times (5 - 2 \sqrt{3} ) \\ = 25 - 10 \sqrt{3} + 10 \sqrt{3} - 4 \times 3 \\ = 25 - 12 \\ = 13$

The required quadratic equation is in the form

${x}^{2} - ( \alpha + \beta )x + \alpha . \beta = 0$

So, by putting the values in the above equation, we get,

${x}^{2} - ( \alpha + \beta )x + \alpha . \beta = 0 \\ {x}^{2} - (10)x + 13 = = 0 \\ \\ {x}^{2} - 10x + 13 = 0$

Ans. :-

The required quadratic equation is

${x}^{2} - 10x + 13 = 0$

Hope it helps!

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