Winter Holiday Homework---- Quadratic Equation #1

$\large\text { To determine whether there exist real roots, the square root of the }$ $\large\text { discriminant in the quadratic formula is used , }$

$\large\displaystyle \sqrt{ b^{2} - 4ac }$

$\large\text { Substitute the values of the coefficient of the quadratic equation, }$ $\large\displaystyle {\sqrt{ (-4)^2 - 4(1)(-9996)} = \sqrt{ 40000 } = 200 }$

$\large\text { Therefore , the quadratic equation has two real roots. }$

$\large\text { To determine the nature of the roots of the quadratic equation , }$ $\large\text { the discriminant is used , }$

$\large\displaystyle { b^{2} - 4ac }$

$\large\text { By substituting the same value into the discriminant, we get }$ $\large\displaystyle { b^{2} - 4ac > 0 } \text{ as } \displaystyle{ 200 > 0 } \text{ which concludes that the quadratic }$ $\large\text { equation to have two distinct real roots. }$

In the quadratic equation $x^2-4x-9996=0,$ $A=1,B=-4$ and $C=-9996$ . We find that

$B^2=(-4)^2=16$

$4AC=4(-9998)=-39984$

Since $B^2>4AC$ , the roots are real and unequal or among the choices: "two different roots".