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Find the value of x. (Be sure to give all solutions)
x^{2}  6 = 138
Simplifying x^{2} + 6 = 138
Reorder the terms: 6 + x^{2} = 138
Solving 6 + x^{2} = 138
Solving for variable 'x'
Move all terms containing x to the left, all other terms to the right
Add '6' to each side of the equation
6 + 6 + x^{2} = 138 + 6
Combine like terms: 6 + 6 = 0
0 + x^{2} = 138 + 6
x^{2} = 138 + 6
Combine like terms: 138 + 6 = 144
x^{2} = 144
Simplifying x^{2} = 144
Take the square root of each side:
x = {12, 12}
Find the values for x for the following equation.
x^{2} + 4x  32 = 0
This equation is written in standard form and it is set equal to 0. Therefore, I know that the 2 methods for solving are factoring and using the quadratic equation.
As I browse the answers, it looks like the values for x are integers, so I must be able to factor. Therefore, this will be the easiest method.
I need to find two integers whose product is 32 and whose sum is 4. (8 & 4).
Don’t make the mistake of assuming these are your answers!!!
(x + 8) (x  4) = 0 Rewrite as factors
x + 8 = 0 x  4 = 0
Set each factor equal to 0
x + 8  8 = 0  8
x  4 + 4 = 0 + 4
x =  8, x = 4
Find the values of y for the following equation:
2y^{2}  5y + 2 = 5
In order to find the values for y, we must set the equation equal to 0 by subtracting 5 from both sides.
2y^{2}  5y + 2  5 = 5  5
Subtract 5 from both sides
2y^{2}  5y  3 = 0
Now the equation is set equal to 0.
As I take a look at the answers, it looks like this equation can be factored. Let’s try!
(2y + 1) (y  3) = 0 Factor the equation
2y + 1 = 0 y  3 = 0
Set each factor equal to 0
2y + 1  1 = 0  1 y  3 + 3 = 0 + 3
2y = 1 y = 3
A ball is shot from a cannon into the air with an upward velocity of 36 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t) = 16t^{2} + 36t + 1.5. Find the maximum height attained by the ball.
In order to find the maximum height of the ball, I would need to find the ycoordinate of the vertex.
Vertex formula: x =  b/2a
where: a = 16 b = 36 c = 1.5
x = 1.125
Now substitute 1.125 for t into the equation and solve for h(t). H(t) = 16 (1.125)^{2} + 36 (1.125) + 1.5
H(t) = 21.75
If a < 0, then function f(x) = ax^{2} + bx + c has a maximum value at
The graph of a quadratic equation is parabola.
If a < 0, then the parabola opens downward. If the parabola opens downward, then the vertex is the point whose yvalue is the maximum value of f.
Quadratic function is a function that can be described by an equation of the form fxx = ax^{2} + bx + c, where a ≠ 0.
In a quadratic function, the greatest power of the variable is 2. The graph of a quadratic function is a parabola.
If α and β are the roots of 3x^{2} + 8x +2 = 0 then what is the value α^{2} + β^{2}?
Given equation is
3x^{2} + 8x + 2 = 0
and α, β are its roots then
If the sum of roots of the equation Kx^{2} + 2x + 3K = 0 is equal to their product, then the value of K is
Here Kx^{2} + 2x + 3K = 0
then given Sum of roots = Product of roots
The roots of a quadratic equation are 7 and 3, then what is the equation?
The required equation is
x^{2}  (Sum of roots)x + Product of roots = 0
⇒ x^{2}  [7 + ( 3)]x + (7) ( 3) = 0
⇒ x^{2}  4x  21 = 0
If one root of the equation x^{2} + px + 12 = 0 is 4, while the equation x^{2} + px + q = 0 has equal roots then the value of q is
Here x^{2} + px + q = 0 has one root a = 4
∴ 42 + b^{4} + 12 = 0
⇒ b = 7
Now the equation x^{2}  7x + q = 0 has equal roots
(7)^{2}  4 (1) (q) = 0 then
⇒ 4q = 49 ⇒ q = 49/4
If α , β are the roots of the equation x^{2}  p (x + 1)  c = 0 then (α + 1) (β + 1) is equal to
The given equation is
x^{2}  p(x + 1)  c = 0
⇒ x^{2}  px + (c  p) = 0
∴ α + β = p
and
Now (α + 1) (β+ 1) = αβ + α + β + 1 =  c p + p + 1 = 1  c
If the difference of the root x^{2} px + q = 0 is unity then
We have x^{2}  px + q = 0
Now α + β = p, αβ = q
∴ α  β = 1
2α = p + 1
Now
β = p  α
and αβ = q
p^{2}  1 = 4q ⇒ p^{2}  4q = 1
If a, b are the roots of the equation x^{2} + x + 1 = 0 then a^{2} + b^{2} = ?
Given a and b are roots of x^{2} + x + 1 = 0
then a + b = 1 and ab = 1
∴ a^{2} + b^{2} = (a + b)^{2}  2ab
= (1)^{2}  2(1) = 1  2 = 1
If A and B are the roots of the quadratic equation x^{2}  12x + 27 = 0, then A^{3} + B^{3} is
Given A and B are the roots of x^{2}  12x + 27 = 0
⇒ (x  3)(x  9) = 0
∴ A = 3, B = 9
Hence A^{3} + B^{3 }= 3^{3} + 9^{3} = 27 + 729 = 756
If the quadratic equation (a^{2}  b^{2})x^{2} + (b^{2}  c^{2})x + (c^{2}  a^{2}) = 0 has equal roots, then which of the following is true?
If the quadratic equation is given by Ax^{2} + Bx + C = 0.
Then,
Given: (a^{2}  b^{2})x^{2} + (b^{2}  c^{2})x + (c^{2}  a^{2}) = 0
Where,
A = (a^{2}  b^{2}), B = (b^{2}  c^{2}), C = c^{2}  a^{2}
Calculation:
Here,
A + B + C = (a^{2}  b^{2}) + (b^{2}  c^{2}) + (c^{2}  a^{2}) = 0.
So, we can say that x = 1 is a root of a given quadratic equation. According to question both roots are same.
Then,
Roots (x) = 1, 1
(a^{2}  b^{2}) = (c^{2 } a^{2})
By arranging, we get
b^{2} + c^{2} = 2a^{2}
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