How do you solve the quadratic equation #5x^2 + 3x = 1# ?

2 Answers
Aug 5, 2016

Answer:

#x_(1,2) = (-5 +- sqrt(13))/10#

Explanation:

The first thing to do here is rearrange your equation to quadratic form by getting all the terms on one side of the equation.

To do that, subtract #1# from both sides of the equation

#5x^2 + 3x - 1 = color(red)(cancel(color(black)(1))) - color(red)(cancel(color(black)(1)))#

#5x^2 + 3x - 1 = 0#

Now, the quadratic formula allows you to calculate the two solutions of a general form quadratic equation

#color(blue)(ax^2 + bx + c = 0)#

by using the equation

#color(blue)(|bar(ul(color(white)(a/a)x_(1,2) = (- b +- sqrt(b^2 - 4 * a * c))/(2 * a) color(white)(a/a)|)))#

In your case, you have

#{(a = color(white)(-)5), (b = color(white)(-)3), (c = -1) :}#

Plug these values into the quadratic formula to get

#x_(1,2) = (-5 +- sqrt(3^2 - 4 * 1 * (-1)))/(2 * 5)#

#x_(1,2) = (-5 +- sqrt(13))/10 implies {(x_1 = (-5 - sqrt(13))/10), (x_2 = (-5 + sqrt(13))/10) :}#

You can thus say that your original equation has two solutions

#x = (-5 - sqrt(13))/10" "# and #" "x = (-5 + sqrt(13))/10#

graph{5x^2 + 3x - 1 [-3, 3, -3, 3]}

Aug 5, 2016

Answer:

#(-3 +- sqrt29)/10#

Explanation:

Bring the quadratic equation to standard form:
#y = 5x^2 + 3x - 1 = 0#
Use the new quadratic formula in graphic form:
#D = d^2 = b^2 - 4ac = 9 + 20 = 29# --> #d = +- sqrt29#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = -3/10 +- sqrt29/10 = (-3 +- sqrt29)/10#

NOTE. Using the improved quadratic formula gets simpler expressions and easier numeric computation. In addition, it shows students the graphic representation and interpretation of the axis of symmetry and the 2 x-intercepts.
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