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Math Olympiad Test: Quadratic Equations- 1


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10 Questions MCQ Test International Mathematics Olympiad (IMO) for Class 10 | Math Olympiad Test: Quadratic Equations- 1

Math Olympiad Test: Quadratic Equations- 1 for Class 10 2022 is part of International Mathematics Olympiad (IMO) for Class 10 preparation. The Math Olympiad Test: Quadratic Equations- 1 questions and answers have been prepared according to the Class 10 exam syllabus.The Math Olympiad Test: Quadratic Equations- 1 MCQs are made for Class 10 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Math Olympiad Test: Quadratic Equations- 1 below.
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Math Olympiad Test: Quadratic Equations- 1 - Question 1

The area of a square is 169 cm2. What is the length of one side of the square?

Detailed Solution for Math Olympiad Test: Quadratic Equations- 1 - Question 1

Since a square has sides with the same length, the formula for Area is A = s2
Let’s substitute 169 for the Area.
s2 = A
s2 = 169
s = √169
s = 13
In this situation, -13 doesn’t apply since the side of a square can’t be a negative number.

Math Olympiad Test: Quadratic Equations- 1 - Question 2

A right triangle has a side with length 12 in and a hypotenuse with length 20 in. Find the length of the second leg. (Round to the nearest hundredth if needed)

Detailed Solution for Math Olympiad Test: Quadratic Equations- 1 - Question 2

Since we are working with a right triangle, and we are missing a leg, we can use the Pythagorean Theorem to solve for that missing leg
Pythagorean Theorem: A2 + B2 = C2
A = 12, B = ?, C = 20
122 + B2 = 202 Substitute
144 + B2 = 400 Simplify
144 - 144 + B2 = 400 - 144
Subtract 144 from both sides
B2 = 256 Simplify
Take the square root of both sides
B = 16

Math Olympiad Test: Quadratic Equations- 1 - Question 3

Find the x-intercepts for the following equation.
Y =  6x2 - 7x - 3

Detailed Solution for Math Olympiad Test: Quadratic Equations- 1 - Question 3

In order to find the x-intercepts, we can let y = 0 and solve for x. We can solve by possible factoring or by using the quadratic formula. Let’s use the discriminant to decide which method would be best.
0 = 6x2 - 7x - 3
where a = 6
b = -7
c = -3
Discriminate: b2 - 4ac
(-7)2 - 4(6)(-3) = 121.
This is positive and a perfect square, so let’s factor!

Math Olympiad Test: Quadratic Equations- 1 - Question 4

Use the quadratic formula to find the values of x for the equation:
x2 - 4x - 10 = 0

Detailed Solution for Math Olympiad Test: Quadratic Equations- 1 - Question 4

Given a quadratic equation of the form
ax2 + bx + c = 0
the quadratic roots can be evaluated using the formula

For x2 − 4x + 10 = 0
∴ x = 5.74 and x = -1.74

Math Olympiad Test: Quadratic Equations- 1 - Question 5

Which statement best describes the solutions to the equation below?
3x2 - 5x + 20 = 0

Detailed Solution for Math Olympiad Test: Quadratic Equations- 1 - Question 5

In order to determine which statement best represents this equation, I would need to find the discriminate. Discriminate: b2 - 4ac
where: a = 3 b = -5 c = 20
(-5)2 - 4(3) (20) = -215
Since the discriminate is negative, this means that there are no real solutions.

Math Olympiad Test: Quadratic Equations- 1 - Question 6

Solution set of 6x2 + x - 15 = 0 is

Detailed Solution for Math Olympiad Test: Quadratic Equations- 1 - Question 6

Math Olympiad Test: Quadratic Equations- 1 - Question 7

Graph of quadratic function is

Detailed Solution for Math Olympiad Test: Quadratic Equations- 1 - Question 7

The graph of a quadratic function is called a parabola.

Math Olympiad Test: Quadratic Equations- 1 - Question 8

If -5, is a root of the quadratic equation 2x2 + Px - 15 = 0 and the quadratic equation P (x2 +x) + K = 0 has equal roots. What is the value of K?

Detailed Solution for Math Olympiad Test: Quadratic Equations- 1 - Question 8

The given equation is
2x2 + Px - 15 = 0
2(-5)2 + P(-5) - 15 = 0
⇒ 50 - 5P - 15 = 0
⇒ -5P = - 35
⇒ P = 7
P(x2 + x) + K = 0
⇒ 7(x2 + x) + K = 0 ⇒ 7x2 + 7x + K = 0
⇒ a = 7, b = 7, c = K
∴ D = b2 - 4ac = (7)2 - 4(7)K = 49 - 28K
The equation has equal roots
D = 0 ⇒ 49 - 28K = 0 ⇒ K = 49/28 = 7/4

Math Olympiad Test: Quadratic Equations- 1 - Question 9

If one root of 3x2 + 11x +K = 0 is reciprocal of the other then what is the value of K?

Detailed Solution for Math Olympiad Test: Quadratic Equations- 1 - Question 9

The given equation is
3x2 + 11x + K = 0
Let one root be α then other root is 1/α

Math Olympiad Test: Quadratic Equations- 1 - Question 10

If α, β are the roots of the equation 3x2 + 8x + 2 = 0 then the value of (1/α + 1/β) is

Detailed Solution for Math Olympiad Test: Quadratic Equations- 1 - Question 10

We have 3x2 + 8x + 2 = 0
and α + β = -8/3, αβ = 2/3
Now

 

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