Respuesta :
Answer:
1. x^2 − 4x + 3 = 0
[tex] b^2 -4ac = (-4)^2 -4(1)*(3)= 4 >0[/tex] So we have two real solutions
2. 2n^2 + 7 = −4n + 5
[tex] b^2 -4ac = (4)^2 -4(2)*(2)= 0 [/tex] So we just one real solution
3. x − 3x^2 = 5 + 2x − x^2
[tex] b^2 -4ac = (1)^2 -4(2)*(5)= -19 <0 [/tex] No real solutions
4. 4x + 7 = x^2 − 5x + 1
[tex] b^2 -4ac = (-9)^2 -4(1)*(-6)= 105 >0 [/tex] So we have two real solutions
Step-by-step explanation:
1. x^2 − 4x + 3 = 0
We need to compare this function with the general equation for a quadratic formula given by:
[tex] f(x) = ax^2 + bx + c[/tex]
On this case we see that a=1, b = -4 and c =3
We can find the discriminat with the following formula:
[tex] \sqrt{b^2 -4ac}[/tex]
[tex] b^2 -4ac = (-4)^2 -4(1)*(3)= 4 >0[/tex] So we have two real solutions
2. 2n^2 + 7 = −4n + 5
We can rewrite the expression like this:
[tex] 2n^2 +4n +2[/tex]
On this case we see that a=2, b = 4 and c =2
We can find the discriminat with the following formula:
[tex] \sqrt{b^2 -4ac}[/tex]
[tex] b^2 -4ac = (4)^2 -4(2)*(2)= 0 [/tex] So we just one real solution
3. x − 3x^2 = 5 + 2x − x^2
We can rewrite the expression like this:
[tex] 2x^2 +x +5[/tex]
On this case we see that a=2, b = 1 and c =5
We can find the discriminat with the following formula:
[tex] \sqrt{b^2 -4ac}[/tex]
[tex] b^2 -4ac = (1)^2 -4(2)*(5)= -19 <0 [/tex] No real solutions
4. 4x + 7 = x^2 − 5x + 1
We can rewrite the expression like this:
[tex] x^2 -9x -6[/tex]
On this case we see that a=1, b = -9 and c =-6
We can find the discriminat with the following formula:
[tex] \sqrt{b^2 -4ac}[/tex]
[tex] b^2 -4ac = (-9)^2 -4(1)*(-6)= 105 >0 [/tex] So we have two real solutions
