Can someone help me solve this: Decide all values of b in the following equations that will give one or more real number solutions.

(a) 3x^2+bx-3=0
(b) 5x^2+bx+1=0
(c) -3x^2 +bx-3=0
(d) write a rule for judging if an equation has solutions by looking at it in standard form.

look at the value of the discriminat

b^2 - 4ac from the quadratic equation formula

if the Discriminant is positive, there are two real and distinct roots
If the Discriminant is zero, there is one real root
if the Discriminant is negative there are two complex roots (imaginary roots)

if the Discriminant is a perfect square ,then there are two rational roots

clearly if a and c have different signs, there will be real solutions.

if a and c have the same sign (either both positive or both negative) then
b^2 > │ac│

look at the third one
b^2 - 4(-3)(-3) > 0 to have real roots
b^2 - 36 > 0
b^2 > 36

so b > 6 or b < -6

a) 3x² + bx - 3 = 0 b) 5x² + bx + 1 = 0 c) -3x² + bx - 3 = 0

bx = -3x² + 3 bx = -5x² - 1 bx = 3x² + 3

bx/x = -3x² + 3/x bx/x = 5x² -1/x bx/x = 3x² + x3/

b = 3(-x² + 1)/x b = - 5x² + 1/x b = 3(x² + 1)/x

b = 3(1² - x²)/x

b = 3[(1 - x)(x + x)]/x

b = 3( -x + 1)(x+1)/x

To find the values of b that will give one or more real number solutions, we need to determine the discriminant of each equation. The discriminant, denoted as Δ (delta), is a mathematical term that can be calculated using the coefficients of a quadratic equation in the form ax^2 + bx + c = 0.

For the quadratic equation ax^2 + bx + c = 0, the discriminant is given by the formula Δ = b^2 - 4ac. We can then use the value of Δ to determine the nature of the solutions:

1. If Δ > 0, the equation has two distinct real solutions.
2. If Δ = 0, the equation has one real solution (which is a repeated root or double root).
3. If Δ < 0, the equation has no real solutions (the solutions are complex or imaginary).

Now let's apply this method to each equation:

(a) 3x^2 + bx - 3 = 0
Here, a = 3, b = b, and c = -3.
The discriminant can be calculated as Δ = b^2 - 4ac.
So, Δ = b^2 - 4(3)(-3) = b^2 + 36.

In order for this equation to have one or more real number solutions, Δ ≥ 0.
Hence, b^2 + 36 ≥ 0.
Subtracting 36 from both sides, we get b^2 ≥ -36.
Since b^2 is always positive or zero, this inequality is satisfied for all real values of b. Therefore, any value of b will give one or more real number solutions to this equation.

(b) 5x^2 + bx + 1 = 0
Here, a = 5, b = b, and c = 1.
The discriminant can be calculated as Δ = b^2 - 4ac.
So, Δ = b^2 - 4(5)(1) = b^2 - 20.

In order for this equation to have one or more real number solutions, Δ ≥ 0.
Hence, b^2 - 20 ≥ 0.
Adding 20 to both sides, we get b^2 ≥ 20.
Taking the square root of both sides, we have |b| ≥ √20.
Therefore, b ≥ √20 or b ≤ -√20.
To simplify this further, we can write b ≥ 2√5 or b ≤ -2√5.
So, any value of b greater than or equal to 2√5 or less than or equal to -2√5 will give one or more real number solutions to this equation.

(c) -3x^2 + bx - 3 = 0
Here, a = -3, b = b, and c = -3.
The discriminant can be calculated as Δ = b^2 - 4ac.
So, Δ = b^2 - 4(-3)(-3) = b^2 - 36.

In order for this equation to have one or more real number solutions, Δ ≥ 0.
Hence, b^2 - 36 ≥ 0.
Adding 36 to both sides, we get b^2 ≥ 36.
Taking the square root of both sides, we have |b| ≥ 6.
Therefore, b ≥ 6 or b ≤ -6.
So, any value of b greater than or equal to 6 or less than or equal to -6 will give one or more real number solutions to this equation.

(d) The general rule for judging if a quadratic equation has real solutions by looking at it in standard form (ax^2 + bx + c = 0) is as follows:

If the discriminant Δ = b^2 - 4ac is greater than or equal to zero (Δ ≥ 0), then the equation has one or more real number solutions.
If the discriminant Δ = b^2 - 4ac is less than zero (Δ < 0), then the equation has no real number solutions.

By using the discriminant, we can quickly determine if a quadratic equation has real solutions without having to actually solve the equation.