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

May 6, 2018

$\triangle > 0 \implies$$\text{Both the roots are "color(red)"REAL and Different}$

#### Explanation:

Here,

$y = f \left(x\right) = 5 {x}^{2} - 13 x - 4 - x \left(x - 1\right)$

$f \left(x\right) = 5 {x}^{2} - 13 x - 4 - {x}^{2} + x$

$f \left(x\right) = 4 {x}^{2} - 12 x - 4$

We have to represents $\left(4 {x}^{2} - 12 x - 4\right)$ as a quadratic

Let, $4 {x}^{2} - 12 x - 4 = 0$

Comparing with $a {x}^{2} + b x + c = 0 ,$ we get

$a = 4 , b = - 12 \mathmr{and} c = - 4$

Now, $\triangle = {b}^{2} - 4 a c = 144 - 4 \left(4\right) \left(- 4\right) = 208 > 0$

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" $\textcolor{w h i t e}{, \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots}$.$\text{color(blue)"conjugate complex root}$

May 6, 2018

Roots are real and $x = \frac{3}{2} \pm \frac{\sqrt{13}}{2}$

#### Explanation:

$y = 5 {x}^{2} - 13 x - 4 - x \left(x - 1\right)$ or

$y = 5 {x}^{2} - 13 x - 4 - {x}^{2} + x$ or

$y = 4 {x}^{2} - 12 x - 4$ or

$y = 4 \left({x}^{2} - 3 x - 1\right)$ 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: $x = \frac{- b \pm \sqrt{D}}{2 a}$or

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

$y = 4 \left(x - \left(\frac{3}{2} + \frac{\sqrt{13}}{2}\right)\right) \left(x - \left(\frac{3}{2} - \frac{\sqrt{13}}{2}\right)\right)$

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

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