How do you find the discriminant of $5 x^{2} - 4 x + 1 = 3 x$ $5x^2-4x+1=3x$ and use it to determine if the equation has one, two real or two imaginary roots?

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How do you find the discriminant of $5 x^{2} - 4 x + 1 = 3 x$ $5x^2-4x+1=3x$ and use it to determine if the equation has one, two real or two imaginary roots?

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Answer 1 · 2017-10-02T16:49:45

See a solution process below:

Explanation:

First, we need to put the equation in standard form. Subtract $3 x$ $color(red)(3x)$ from each side of the equation to put the equation in standard form while keeping the equation balanced:

$5 x^{2} - 4 x - 3 x + 1 = 3 x - 3 x$ $5x^2 - 4x - color(red)(3x) + 1 = 3x - color(red)(3x)$

$5 x^{2} + (- 4 - 3) x + 1 = 0$ $5x^2 + (-4 - color(red)(3))x + 1 = 0$

$5 x^{2} + (- 7) x + 1 = 0$ $5x^2 + (-7)x + 1 = 0$

$5 x^{2} - 7 x + 1 = 0$ $5x^2 - 7x + 1 = 0$

The quadratic formula states:

For $a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$ , the values of $x$ $x$ which are the solutions to the equation are given by:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2 - 4ac))/(2a)$

The discriminate is the portion of the quadratic equation within the radical: $b^{2} - 4 a c$ $color(blue)(b)^2 - 4color(red)(a)color(green)(c)$

If the discriminate is:
- Positive, you will get two real solutions
- Zero you get just ONE solution
- Negative you get complex solutions

To find the discriminant for this problem substitute:

$5$ $color(red)(5)$ for $a$ $color(red)(a)$

$- 7$ $color(blue)(-7)$ for $b$ $color(blue)(b)$

$1$ $color(green)(1)$ for $c$ $color(green)(c)$

${(- 7)}^{2} - (4 \cdot 5 \cdot 1)$ $color(blue)((-7))^2 - (4 * color(red)(5) * color(green)(1))$

$49 - 20$ $49 - 20$

$29$ $29$

Because the discriminate is positive, you will get two real solutions or roots.

Answer 2 · 2017-10-02T16:54:48

See below.

Explanation:

Arrange $5 x^{2} - 4 x + 1 = 3 x$ $5x^2-4x+1=3x$

To get: $5 x^{2} - 7 x + 1 = 0$ $5x^2-7x+1=0$

We now have the form:

$a x^{2} + b x + 2$ $ax^2+bx+2$

The discriminant of a quadratic is:

$\sqrt{b^{2} - 4 a c}$ $sqrt(b^2-4ac)$

if: $b^{2} - 4 a c > 0$ $b^2-4ac > 0$ then the roots are real and different.

if: $b^{2} - 4 a c = 0$ $b^2-4ac =0$ then the roots are real and repeated,( this is sometimes just called a single root).

if: $b^{2} - 4 a c < 0$ $b^2-4ac < 0$ then the roots are imaginary ( we will have the root of a negative number).

Hope this helps.

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How do you find the discriminant of $5 x^{2} - 4 x + 1 = 3 x$ $5x^2-4x+1=3x$ and use it to determine if the equation has one, two real or two imaginary roots?

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How do you find the discriminant of 5x^2-4x+1=3x5x2−4x+1=3x5x^2-4x+1=3x and use it to determine if the equation has one, two real or two imaginary roots?

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Explanation:

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How do you find the discriminant of $5 x^{2} - 4 x + 1 = 3 x$ $5x^2-4x+1=3x$ and use it to determine if the equation has one, two real or two imaginary roots?