Help with algebra???

Question

can someone help me with these four???

Find the x-intercepts of the graph of the equation:

1. y = x^2 - 3x +2
2. y = x^2 - 1

Decide how many solutions the equation has

1. x^2 - 2x + 1 = 0
2. x^2 + 3 = 0

thank you!

Answer 1

1. x-intercepts- when it crosses the x axis, or when y = 0
0 = x^2 - 3x + 2 factor and solve (x-2)(x-1) the ans 2,0 and 1,0

2. x^2 = 1 x = plus or minus 1 1,0 and -1,0

solutions 1 factor (x-1)(x-1) one solution x = 1
2. x^2 = -3 x = +- sqrt(-3) no real solutions, two imaginary solutions. You will have to decide which answer your teacher wants. (They are both correct)

Answer 2

The x-intercept is where the graph crosses the x axis. What is the y coordinate of any point on the x axis? Zero, right?

Set each of the equations to zero.
1. x² - 3x +2 = 0
2. x² – 1= 0

Now, you can solve them by factoring, or by using the quadratic formula.


There are always 2 solutions to a quadratic equation. However sometimes the solutions are equal, and sometimes they are outside the set of real numbers. I presume you are talking about real solutions.
1. x² - 2x + 1 = 0
2. x² + 3 = 0

You can factor the first one. The second one you’ll have to use the quadratic formula.

Answer 3

1. y = x^2 - 3x +2
When x = 0, y=2 so y-intercept is (0,2)
When y = 0, x = 1 or 2 So (1,0 and (2,0) are x-intercepts

2. y = x^2 - 1
When x = 0, y = -1 so (0,-1) is the y-intercept
When y = 0 x = 1 or -1 So (1,0) and (-1,0) are x-intercepts.

1. x^2 - 2x + 1 = 0
Since b^2-4ac = 0, there is a double root =1. So you could say there is just 1 root since the two roots equal each other.

2. x^2 + 3 = 0
b^2-4ac = -12. Thus there are no real roots. The two roots are imaginary.

Note in the equation ax^2 +bx +c, if b^2-4ac = 0 there is one root, if b^-4ac >0, there are two real roots, if b^2-4ac <0 there are no real roots.

Answer 4

for the first one you have y=0 so #1 X= -2 and -1 #2 X= -1 and 1

#3 is two solutions
and #4 has one solution

Answer 5

Help with algebra???
can someone help me with these four???

Find the x-intercepts of the graph of the equation:puty=0

1. y = x^2 - 3x +2,x=1,2
2. y = x^2 - 1x=+/-1
1. x^2 - 2x + 1 = 0 x-intercepts=1
2. x^2 + 3 = 0, x-intercept=none, imaginary.no intersection with the x-axis.

Answer 6

For the first two, just use a graphing calculator, very simple, "y=" enter those two, then you can find the intercept of both. (2nd trace) Which is, X=1, y=0

Answer 7

1. y=0
x^2-3x+2 = (x-1)(x-2)=> x1=1; x2=2

2.y=0
x^2-1=(x-1)(x+1)=> x1=1; x2=-1

1. x^2-2x+1=(x-1)^2=0 =>x1=x2=1...one solution
2. x^2=-3 ...no solution when working with real numbers