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
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.
2 answers
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
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
3 answers