Grade 10 Math Question

Question

Determine the number of roots, using the most efficient method possible. Please show all work.

a) y = -5(x + 1)² - 3 

b) y = 2(x - 1)(x + 3)

c) y = 2x² + 3x + 1

Answer 1

all quadratic equations have 2 solutions or roots, either both real or both imaginary

the two real might be repetitive (the same value), if the discriminant = 0

the discriminant is b^2-4ac for a quadratic equation in the form y=ax^2 + bx + c

a) y = -5(x+1)^2 - 3

=-5(x^2 +2x +1) -3

=-5x^2 -10x -8

the discriminant = b^2 -4ac = 100-160 <0, so y has 2 imaginary solutions

using the quadratic formula

x = 10/-10 + or - (1/10)sqr(-60)= about -1 + 0.775i

imaginary solutions come in conjugate pairs

when the discriminant is negative, there are two imaginary solutions

or just looking at the equation y =-5(x+1)^2 -3, you can read off the vertex & maximum point (-1,-3) and know it's a downward opening parabola (given the negative 5 coefficient for the x^2 term) that never intesects the x axis, so it has no real solutions.

b) y= 2(x-1)(x+3) has 2 real solutions or roots: x =1 or -3, just set each factor =0 and solve for x

c) y=2x^2 + 3x + 1 factor to get

=(2x+1)(x+1) = 0

2x+1 = 0 or x+1 = 0

x = -1/2, or -1. Two real solutions or roots

you can use "solution," "root," "zero" or "x intercept" nearly interchangeably for the value of x when the equation is solved for y