How do you solve $10 x^{2} - 31 x + 15 = 0$ $10x^2-31x+15=0$ using the quadratic formula?

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How do you solve $10 x^{2} - 31 x + 15 = 0$ $10x^2-31x+15=0$ using the quadratic formula?

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Answer 1 · 2017-06-12T13:26:11

See a solution process below:

Explanation:

From: http://www.purplemath.com/modules/quadform.htm

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:

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

Substituting $10$ $10$ for $a$ $a$ ; $- 31$ $-31$ for $b$ $b$ and $15$ $15$ for $c$ $c$ gives:

$x = \frac{- (- 31) \pm \sqrt{{(- 31)}^{2} - (4 \cdot 10 \cdot 15)}}{2 \cdot 10}$ $x = (-(-31) +- sqrt((-31)^2 - (4 * 10 * 15)))/(2 * 10)$

$x = \frac{31 \pm \sqrt{961 - 600}}{20}$ $x = (31 +- sqrt(961 - 600))/20$

$x = \frac{31 \pm \sqrt{361}}{20}$ $x = (31 +- sqrt(361))/20$

$x = \frac{31 \pm 19}{20}$ $x = (31 +- 19)/20$

$x = \frac{50}{20}$ $x = 50/20$ and $x = \frac{12}{20}$ $x = 12/20$

$x = \frac{5}{2}$ $x = 5/2$ and $x = \frac{3}{5}$ $x = 3/5$

Answer 2 · 2017-06-12T16:46:37

$\frac{5}{2}, \frac{6}{5}$ $5/2, 6/5$

Explanation:

Use the improved quadratic formula (Google, Socratic Search).
$f (x) = 10 x^{2} - 31 x + 15 = 0$ $f(x) = 10x^2 - 31x + 15 = 0$
$D = d^{2} = b^{2} - 4 a c = 961 - 600 = 361$ $D = d^2 = b^2 - 4ac = 961 - 600 = 361$ --> $d = \pm 19$ $d = +- 19$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{31}{20} \pm \frac{19}{20} = \frac{31 \pm 19}{20}$ $x = -b/(2a) +- d/(2a) = 31/20 +- 19/20 = (31 +- 19)/20$
$x 1 = \frac{50}{20} = \frac{5}{2}$ $x1 = 50/20 = 5/2$
$x 2 = \frac{12}{20} = \frac{3}{5}$ $x2 = 12/20 = 3/5$

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How do you solve $10 x^{2} - 31 x + 15 = 0$ $10x^2-31x+15=0$ using the quadratic formula?

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