Question

Without solving, determine the number of real solutions for each quadratic equation.

Answer 1

Answer:

1. x^2 − 4x + 3 = 0

$b^2 -4ac = (-4)^2 -4(1)*(3)= 4 >0$ So we have two real solutions

2. 2n^2 + 7 = −4n + 5

$b^2 -4ac = (4)^2 -4(2)*(2)= 0$ So we just one real solution

3. x − 3x^2 = 5 + 2x − x^2

$b^2 -4ac = (1)^2 -4(2)*(5)= -19$ No real solutions

4. 4x + 7 = x^2 − 5x + 1

$b^2 -4ac = (-9)^2 -4(1)*(-6)= 105 >0$ 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:

$f(x) = ax^2 + bx + c$

On this case we see that a=1, b = -4 and c =3

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (-4)^2 -4(1)*(3)= 4 >0$ So we have two real solutions

2. 2n^2 + 7 = −4n + 5

We can rewrite the expression like this:

$2n^2 +4n +2$

On this case we see that a=2, b = 4 and c =2

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (4)^2 -4(2)*(2)= 0$ So we just one real solution

3. x − 3x^2 = 5 + 2x − x^2

We can rewrite the expression like this:

$2x^2 +x +5$

On this case we see that a=2, b = 1 and c =5

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (1)^2 -4(2)*(5)= -19$ No real solutions

4. 4x + 7 = x^2 − 5x + 1

We can rewrite the expression like this:

$x^2 -9x -6$

On this case we see that a=1, b = -9 and c =-6

We can find the discriminat with the following formula:

$\sqrt{b^2 -4ac}$

$b^2 -4ac = (-9)^2 -4(1)*(-6)= 105 >0$ So we have two real solutions