1. The quadratic equation -3x^2 – x = 10 has two real solutions. (true or False) Justify your answer.
2. If (2x – 3) (x + 5) = 8, then either 2x – 3 = 8 or x + 5 = 8. (True or False) Justify your answer.
-3x^2 – x = 10
3x^2 + x + 10=0
the discriminant is
b^2 - 4ac
= 1 - 4(3)(1) which is negative, so
there are two complex solutions, and the statement is false
2. again it is false,
it would be true if the right side = 0 and not 8
or take one the answers,
x+5 = 8
x = 3
sub that into the original
LS = (6-3)(3+5)
= 3(8) = 24
RS = 8
LS ≠ RS, the statement is false
Thank you so much Reiny. I clearly understood it.
1. The quadratic equation -3x^2 – x = 10 has two real solutions. (False)
To solve a quadratic equation, we can set it equal to zero and apply the quadratic formula. The quadratic formula is given as:
x = (-b ± √(b^2 - 4ac)) / 2a
For the given equation -3x^2 – x = 10, we can rearrange it to:
-3x^2 – x - 10 = 0
Comparing this equation to the general form ax^2 + bx + c = 0, we have a = -3, b = -1, and c = -10.
Now, substituting these values into the quadratic formula, we get:
x = (-(-1) ± √((-1)^2 - 4(-3)(-10))) / (2*(-3))
x = (1 ± √(1 - 120)) / (-6)
x = (1 ± √(-119)) / (-6)
Since the term √(-119) is imaginary, there are no real solutions to this equation. Therefore, the statement "The quadratic equation -3x^2 – x = 10 has two real solutions" is false.
2. If (2x – 3) (x + 5) = 8, then either 2x – 3 = 8 or x + 5 = 8. (False)
To determine if the statement is true or false, we can expand the given equation and simplify it.
Expanding the equation (2x – 3)(x + 5), we get:
2x^2 + 10x - 3x - 15 = 8
2x^2 + 7x - 15 = 8
Now, rearranging the equation to standard quadratic form, we have:
2x^2 + 7x - 23 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. However, none of these methods will result in the solution x = 8. Therefore, the statement "either 2x – 3 = 8 or x + 5 = 8" is false.