Algebra 1, Please Help! The Quadratic Formula

Question

Use the Quadratic Formula to solve the equation 2x² + 5x − 12 = 0. Show all steps/work.

Answer 1

The quadratic formula is X = ( -b ± √ b² - 4ac)/ 2a. In this case, a = 2, b = 5, and c = -12.

Substituting the numbers for the letters in the formula, we get X = (-5 ± √ 25 - 4 x 2 x -12) / 4.

Evaluating further, X = (-5 ± √ 25 + 96 ) / 4

X = (-5 ± √ 121 ) / 4 = (-5 ± 11) /4 = 1.5, -4 there are two roots.

I hope this is clear.

Answer 2

2x^2 +5x -12 = 0

ax^2 +bx +c =0 is the standard quadratic equation

in this problem a=2, b=5 and c=-12

the quadratic formula is

x = -b/2a + or - (1/2a)sqr(b^2 -4ac)

plug in the numbers and you get

x = -5/4 + or - (1/4)sqr(25 +4(2)12))

x= -5/4 + or - (1/4)sqr(121)

x = -5/4 + or - 11/4

x = -16/4 or 6/4

x = -4 or 1.5

check the answer by factoring the polynomial

2x^2 +5x -12

= (2x-3)(x+4)

set each factor = 0 and solve for x

x =-4, 1.5

or by completing the square

2x^2 +5x-12 =0

x^2 +5x/2 -6 =0

x^2 + 5x/2 + (5/4)^2 - 6 - (5/4)^2 =0

(x +5/4)^2 - 121/16 =0

(x+5/4)^2 = 121/16

x+5/4 = + or - 11/4

x = -5/4 + or - 11/4

x = -16/4 or 6/4

x = -4 or 1.5

or you could graph it and find the x intercepts

it's a parabola, upward opening as a = 2>0

2x^2 +5x -12 has y intercept = -12

x intercepts = -4 and 1.5

the x coordinate of the vertex is the average of the x intercepts (-4+1.5)/2 = -2.5/2 = -1.25

the vertex x coordinate is the -b/2a term of the quadratic formula = -5/4

plug it into the polynomial to find the y coordinate of the vertex

2(5/4)^2 +5(5/4) -12 = -15 1/8

the vertex is (-1.25, -15.125) = (-5/4, -121/8) = (-1 1/4, -15 1/8)

axis of symmetry is x=-5/4 which is the average of the two x intercepts

the x intercepts are the roots or solutions to the quadratic equation.

Just a rough sketch might give a good clue as to the general range of

where the parabola intersects the x axis