$16x^2-8x-3=0$ Let $x=q$ and $x=r$ be solutions to the equation shown, with q>r. What is the value of $q-r$ ?

View step-by-step help

Home

Math Problems

Algebra 1

Write linear equations in standard form

Full solution

16x^2-8x-3=0

Let

x=q

and

x=r

be solutions to the equation shown, with

q>r

. What is the value of

q-r

Calculate Discriminant: To find the value of $q - r$ , we can use the quadratic formula, which states that for any quadratic equation $ax^2 + bx + c = 0$ , the solutions are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . Here, $a = 16$ , $b = -8$ , and $c = -3$ .
Apply Quadratic Formula: First, we calculate the discriminant, which is $b^2 - 4ac$ . Plugging in the values, we get $(-8)^2 - 4(16)(-3)$ .
Simplify Solutions: Calculating the discriminant: $(-8)^2 - 4(16)(-3) = 64 + 192 = 256$ .
Find $q$ and $r$ : Now, we apply the quadratic formula. The two solutions are $x = \frac{-(-8) \pm \sqrt{256}}{2 \times 16}.$
Calculate $q - r$ : Simplifying the solutions, we get $x = (8 \pm \sqrt{256}) / 32$ .
Calculate $q - r$ : Simplifying the solutions, we get $x = (8 \pm \sqrt{256}) / 32$ .Since $\sqrt{256} = 16$ , the solutions are $x = (8 \pm 16) / 32$ .
Calculate q - r: Simplifying the solutions, we get $x = \frac{8 \pm \sqrt{256}}{32}$ .Since $\sqrt{256} = 16$ , the solutions are $x = \frac{8 \pm 16}{32}$ .The two solutions are $x = \frac{8 + 16}{32}$ and $x = \frac{8 - 16}{32}$ , which simplify to $x = \frac{24}{32}$ and $x = \frac{-8}{32}$ .
Calculate q - r: Simplifying the solutions, we get $x = (8 \pm \sqrt{256}) / 32$ .Since $\sqrt{256} = 16$ , the solutions are $x = (8 \pm 16) / 32$ .The two solutions are $x = (8 + 16) / 32$ and $x = (8 - 16) / 32$ , which simplify to $x = 24 / 32$ and $x = -8 / 32$ .Simplifying the fractions, we get $x = 3/4$ and $x = -1/4$ .
Calculate $q - r$ : Simplifying the solutions, we get $x = (8 \pm \sqrt{256}) / 32$ .Since $\sqrt{256} = 16$ , the solutions are $x = (8 \pm 16) / 32$ .The two solutions are $x = (8 + 16) / 32$ and $x = (8 - 16) / 32$ , which simplify to $x = 24 / 32$ and $x = -8 / 32$ .Simplifying the fractions, we get $x = 3/4$ and $x = -1/4$ .Since $x = (8 \pm \sqrt{256}) / 32$ $0$ , we have $x = (8 \pm \sqrt{256}) / 32$ $1$ and $x = (8 \pm \sqrt{256}) / 32$ $2$ .
Calculate $q - r$ : Simplifying the solutions, we get $x = (8 \pm \sqrt{256}) / 32$ .Since $\sqrt{256} = 16$ , the solutions are $x = (8 \pm 16) / 32$ .The two solutions are $x = (8 + 16) / 32$ and $x = (8 - 16) / 32$ , which simplify to $x = 24 / 32$ and $x = -8 / 32$ .Simplifying the fractions, we get $x = 3/4$ and $x = -1/4$ .Since $x = (8 \pm \sqrt{256}) / 32$ $0$ , we have $x = (8 \pm \sqrt{256}) / 32$ $1$ and $x = (8 \pm \sqrt{256}) / 32$ $2$ .To find $q - r$ , we calculate $x = (8 \pm \sqrt{256}) / 32$ $4$ .
Calculate $q - r$ : Simplifying the solutions, we get $x = (8 \pm \sqrt{256}) / 32$ .Since $\sqrt{256} = 16$ , the solutions are $x = (8 \pm 16) / 32$ .The two solutions are $x = (8 + 16) / 32$ and $x = (8 - 16) / 32$ , which simplify to $x = 24 / 32$ and $x = -8 / 32$ .Simplifying the fractions, we get $x = 3/4$ and $x = -1/4$ .Since $x = (8 \pm \sqrt{256}) / 32$ $0$ , we have $x = (8 \pm \sqrt{256}) / 32$ $1$ and $x = (8 \pm \sqrt{256}) / 32$ $2$ .To find $q - r$ , we calculate $x = (8 \pm \sqrt{256}) / 32$ $4$ .Calculating $q - r$ : $x = (8 \pm \sqrt{256}) / 32$ $6$ .