How do you find the roots, real and imaginary, of #y= 5x^2 - 13x -4-x(x-1) # using the quadratic formula?

Question

Avery Alexander · Accepted Answer

#triangle > 0=>##"Both the roots are "color(red)"REAL and Different"#

Here,

#y=f(x)=5x^2-13x-4-x(x-1)#

#f(x)=5x^2-13x-4-x^2+x#

#f(x)=4x^2-12x-4#

We have to represents #(4x^2-12x-4)# as a quadratic equation,before using quadratic formula.

Let, #4x^2-12x-4=0#

Comparing with #ax^2+bx+c=0,# we get

#a=4,b=-12 and c=-4#

Now, #triangle=b^2-4ac=144-4(4)(-4)=208 > 0#

Hence,

#triangle > 0=>"Both the roots are "color(red)"REAL and Different"#

Note:

#(1)triangle > 0=>"Both the roots are "color(red)"REAL and Different"#

#(2)triangle = 0=>"Both the roots are "color(red)"REAL and Equal"#

#(3)triangle < 0=>"Both the roots are "color(red)"Imaginary and Different"# #color(white)(,................................................)#.#""color(blue)"conjugate complex root"#

Kennedy Adams · Answer

Roots are real and #x = 3/2 +- sqrt 13/2#

#y = 5 x^2 -13 x - 4 - x(x-1)# or

#y = 5 x^2 -13 x - 4 - x^2 +x# or

#y = 4 x^2 -12 x - 4 # or

#y = 4( x^2 -3 x - 1) # Comparing with standard quadratic

equation #ax^2+bx+c=0; a=1 ,b=-3 ,c=-1#

Discriminant # D= b^2-4 a c = 9+4=13 #

Discriminant positive, we get two real roots,

Quadratic formula: #x= (-b+-sqrtD)/(2a) #or

#x= (3+-sqrt 13)/2 = 3/2 +- sqrt 13/2#

#y = 4(x- ( 3/2 + sqrt 13/2))(x- ( 3/2 - sqrt 13/2)) #

Zeros are #x = 3/2 +- sqrt 13/2#

[Ans]

Ethan Adkins · Answer

undefined To find the roots of the equation $ y = 5x^2 - 13x - 4 - x(x - 1) $, first, simplify the equation to $ y = 5x^2 - 13x - 4 - x^2 + x $. Then combine like terms to get $ y = 4x^2 - 12x - 4 $. Now, the equation is in the form $ ax^2 + bx + c $, where $ a = 4 $, $ b = -12 $, and $ c = -4 $.

Apply the quadratic formula $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ to find the roots:

$$ x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 4 \cdot (-4)}}{2 \cdot 4} $$

$$ x = \frac{12 \pm \sqrt{144 + 64}}{8} $$

$$ x = \frac{12 \pm \sqrt{208}}{8} $$

$$ x = \frac{12 \pm 4\sqrt{13}}{8} $$

$$ x = \frac{3 \pm \sqrt{13}}{2} $$

Therefore, the roots of the equation are $ x = \frac{3 + \sqrt{13}}{2} $ and $ x = \frac{3 - \sqrt{13}}{2} $.