Question

Which equation has the components of 0 = x2 – 9x – 20 inserted into the quadratic formula correctly? x = x = x = x =

Answer 1

To solve the quadratic equation given by 0=x^2-9x-20, we use the quadratic formula given by:
x=[-b+\- sqrt(b^2-4ac)]/(2a)
where,
a=1,b=-9,c=-20
thus substituting the above values into our formula we get:
x=[9+\-sqrt(9^2-4(-20*1))/(2*1)
x=[9+\-sqrt(161)]/2
x=[9+sqrt161]/2 or x=[9-sqrt161]/2

Answer 2

The equation that has the components of 0 = x² – 9x – 20 inserted into the quadratic formula correctly is $x = \dfrac{-(-9) \pm \sqrt{(-9)^2-4(1)(-20)}}{2(1)}$.

What is the formula of the quadratic equation?

The formula of the quadratic equation helps us to find the roots of a quadratic equation. It is given as,

$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

where a, b, and c are the coefficient of x², x , and 1, respectively in the quadratic equation ax²+bx+c=0.

We know to put the quadratic equation correctly in the formula, we must know about the constants of a quadratic equation, therefore, compare it with the general quadratic equation,

$ax^2 +bx + c = 0\\x^2-9x -20=0$

a = 1, b = -9, and c = -20,

Substitute the values,

$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x = \dfrac{-(-9) \pm \sqrt{(-9)^2-4(1)(-20)}}{2(1)}\\\\x = \dfrac{9 \pm \sqrt{81+80}}{2}\\\\\\x = \dfrac{9 + \sqrt{81+80}}{2}, \dfrac{9 - \sqrt{81+80}}{2}$

Hence, the equation that has the components of 0 = x² – 9x – 20 inserted into the quadratic formula correctly is $x = \dfrac{-(-9) \pm \sqrt{(-9)^2-4(1)(-20)}}{2(1)}$.

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