Math Olympiad Test: Quadratic Equations- 1 - Class 10 MCQ

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.

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

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!

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

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.

The graph of a quadratic function is called a parabola.

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

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

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