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كويز تفاعلي: Quadratic Formula & The Discriminant - Reveal
Comprehensive practice on the Quadratic Formula and the Discriminant. This worksheet covers the definition of the discriminant, using it to determine the number and type of roots (real or imaginary), and applying the full quadratic formula to solve various quadratic equations, including those with complex solutions.
يرجى الانتباه إلى أن المعلم قام بإعداد الأسئلة فقط، ولم يقم بإعداد الإجابات أو الشروحات المرفقة. وقد تم توليد الإجابات باستخدام تقنيات الذكاء الاصطناعي، لذلك قد تتضمن بعض الأخطاء أو عدم الدقة.
The discriminant is
A
ax2 + bx + c
B
b - 4ac
C
b2 - 4ac
D
b2 + 4ac
Explanation
The discriminant of a quadratic equation in the form ax2 + bx + c = 0 is defined by the formula b2 - 4ac .
Determine the value of the discriminant and describe the number and type roots for the following: x2 + 7x + 13
A
101; 2 real roots
B
3; 2 real roots
C
-101; 2 imaginary roots
D
-3; 2 imaginary roots
Explanation
Using the discriminant formula D = b2 - 4ac with a=1, b=7, c=13 , we get D = 72 - 4(1)(13) = 49 - 52 = -3 . Since the discriminant is negative, the equation has two imaginary roots.
Use the quadratic formula to solve 2x2 + 2x - 12 .
A
x = -2, 3
B
x = 2, 3
C
x = 2, -3
D
x = -2, -3
Explanation
Dividing the equation by 2 gives x2 + x - 6 = 0 . Factoring gives (x + 3)(x - 2) = 0 , so the solutions are x = 2 and x = -3 .
If the discriminant is negative, then the quadratic has:
A
1 Real Solution
B
2 Real Solutions
C
Half a Solution
D
2 Complex (imaginary) Solutions
Explanation
When the discriminant b2 - 4ac < 0 , the square root in the quadratic formula results in an imaginary number, leading to two complex solutions.
The quadratic equation can be used to solve quadratic equations that can or cannot be factored.
Explanation
The quadratic formula is a universal method that can solve any quadratic equation, regardless of whether it is factorable over integers or not.
Solve 2p2 - 3p - 3 = 0 using the Quadratic Formula.
A
\(\frac{3 \pm \sqrt{33}}{4}\)
B
\(\frac{-1 \pm \sqrt{13}}{2}\)
C
\{2, 1\}
D
\(\frac{3 \pm \sqrt{17}}{2}\)
Explanation
For 2p2 - 3p - 3 = 0 , a=2, b=-3, c=-3 . The formula gives \(p = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-3)}}{2(2)} = \frac{3 \pm \sqrt{9 + 24}}{4} = \frac{3 \pm \sqrt{33}}{4}\), which corresponds to option A.
Solve using the quadratic formula: 2x2 - 4x + 7 = 0
A
\(x = \frac{2 \pm 2i\sqrt{10}}{2}\)
B
\(x = \frac{4 \pm \sqrt{40}}{4}\)
C
\(x = \frac{2 \pm i\sqrt{10}}{2}\)
D
\(x = \frac{2 \pm 2i\sqrt{40}}{4}\)
Explanation
For 2x2 - 4x + 7 = 0 , \(x = \frac{4 \pm \sqrt{16 - 4(2)(7)}}{4} = \frac{4 \pm \sqrt{-40}}{4} = \frac{4 \pm 2i\sqrt{10}}{4} = \frac{2 \pm i\sqrt{10}}{2}\).
If the discriminant is equal to 0, how many solutions are there?
A
No solutions
B
\(x = 1\ \text{solution}\)
C
\(x = 2\ \text{solutions}\)
D
\(x = \infty\ \text{solutions}\)
Explanation
When the discriminant is zero, the term under the square root in the quadratic formula is zero, resulting in a single real repeated root.
If the discriminant is equal to 4, how many solutions are there?
A
No solutions
B
\(1\ \text{solution}\)
C
\(2\ \text{solutions}\)
D
\(\infty\ \text{solutions}\)
Explanation
Since 4 is a positive number, a positive discriminant indicates that the quadratic equation has two distinct real solutions.
Solve x2 - x - 2 = 0 using the Quadratic Formula.
A
\(x = \frac{-1 \pm \sqrt{10}}{3}\)
B
\(x = 2,\ -1\)
C
\(x = 2,\ -3\)
D
\(x = 1,\ -2\)
Explanation
The equation x2 - x - 2 = 0 can be factored as (x - 2)(x + 1) = 0 , giving solutions x = 2 and x = -1 . This corresponds to option B.
Use the quadratic formula to determine the solutions to the equation 2x2 - 9x - 35 = 0 .
A
\(x = \frac{7}{2},\x = -6\)
B
\(x = -\frac{5}{2},\x = 5\)
C
\(x = -\frac{3}{7},\x = 6\)
D
\(x = -\frac{5}{2},\x = 7\)
Explanation
Using the formula: \(x = \frac{9 \pm \sqrt{81 - 4(2)(-35)}}{4} = \frac{9 \pm \sqrt{81 + 280}}{4} = \frac{9 \pm \sqrt{361}}{4} = \frac{9 \pm 19}{4}\). This gives x = 7 and \(x = -\frac{5}{2}\).
Solve x2 - 5x + 10 = 0 .
A
\(x = \frac{5 - i\sqrt{15}}{2},\ \frac{5 + i\sqrt{15}}{2}\)
B
\(x = \frac{5 - \sqrt{15}}{2},\ \frac{5 + \sqrt{15}}{2}\)
C
\(x = \frac{5 - i\sqrt{65}}{2},\ \frac{5 + i\sqrt{65}}{2}\)
D
\(x = \frac{5 - \sqrt{65}}{2},\ \frac{5 + \sqrt{65}}{2}\)
Explanation
Using the formula: \(x = \frac{5 \pm \sqrt{25 - 4(1)(10)}}{2} = \frac{5 \pm \sqrt{-15}}{2} = \frac{5 \pm i\sqrt{15}}{2}\).
Solve for x: 3x2 - 6x + 6 = 0
A
\(x = 1 \pm i\)
B
\(x = 3,\ -6\)
C
\(x = 4.8,\ 2.7\)
D
Undefined
Explanation
Dividing by 3 gives x2 - 2x + 2 = 0 . Using the formula: \(x = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i\).
Solve 2x2 - 36 = x .
A
\(x = -\frac{9}{2},\ 4\)
B
\(x = \frac{9}{2},\ -4\)
C
\(x = 4,\ -4\)
D
No real solution
Explanation
Rearrange to 2x2 - x - 36 = 0 . Using the formula: \(x = \frac{1 \pm \sqrt{1 - 4(2)(-36)}}{4} = \frac{1 \pm \sqrt{289}}{4} = \frac{1 \pm 17}{4}\). This gives \(x = \frac{9}{2}\) and x = -4 .
Why are quadratic equations set equal to zero?
A
Because zero is an easy number to work with.
B
Because we are trying to find the x-intercepts and that happens when y = 0.
C
Because setting it equal to zero helps us find the y-intercepts.
D
None of the above
Explanation
In coordinate geometry, the x-intercepts of a function occur where the value of the function (y) is zero. Thus, setting the quadratic equation to zero allows us to find these points.
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