Question

Find the quadratic equation having one root :$2\sqrt{3}-4$

• A. $x^2-4\sqrt{3}x-4=0$
• B. $x^2-8x+4=0$
• C. $x^2+4\sqrt{3}x-4=0$
• D. $x^2-8x-4=0$

Answer 1

Answer: (A)

$x^2-4\sqrt{3}x-4=0$

If one root is $2\sqrt{3}-4$ then the other root will be $2\sqrt{3}+4$
Sum of root $=4\sqrt{3}$
Product of roots $=(2\sqrt{3}-4)(2\sqrt{3}+4)=12-16=-4$
The equation whose standard form is $x^2-Sx+P=0$ will be, $x^2-4\sqrt{3}x-4=0$