Solve the following quadratic equation using the perfect square trinomial pattern: x2+x+0.25=0
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The given quadratic equation is: x^2 + x + 0.25 = 0
To solve this equation using the perfect square trinomial pattern, we need to rewrite the equation in the form of (x + a)^2 = 0, where 'a' is a constant.
Let's compare the given equation with the perfect square trinomial pattern:
x^2 + x + 0.25 = (x + a)^2
By comparing the terms, we can see that a = 0.5, since (0.5)^2 = 0.25.
So, we can rewrite the equation as:
x^2 + x + 0.25 = (x + 0.5)^2
Now, we can apply the zero product property, which states that if the product of two factors is equal to zero, then at least one of the factors must be zero.
(x + 0.5)^2 = 0
Taking the square root of both sides:
x + 0.5 = 0
x = -0.5
Therefore, the solution to the quadratic equation x^2 + x + 0.25 = 0 using the perfect square trinomial pattern is x = -0.5.
Ginger was asked to solve 25x^2−196=0 by factoring using the difference of squares pattern. What is/are the correct solution(s)?
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To solve the quadratic equation 25x^2 - 196 = 0 using the difference of squares pattern, we rewrite it as:
(5x)^2 - 14^2 = 0
Now, we can apply the difference of squares formula, which states that a^2 - b^2 = (a + b)(a - b):
(5x + 14)(5x - 14) = 0
Now, set each factor equal to zero and solve for x:
5x + 14 = 0
5x = -14
x = -14/5
and
5x - 14 = 0
5x = 14
x = 14/5
Therefore, the correct solutions to the quadratic equation 25x^2 - 196 = 0 using the difference of squares pattern are x = -14/5 and x = 14/5.
Which of the following is a solution to the quadratic equation −x2+x=−10 ? Assume that the solution has been rounded to the nearest hundredth, if applicable. Show your work
To find the solution to the quadratic equation -x^2 + x = -10, we rearrange the equation to the standard form:
x^2 - x + 10 = 0
We can solve this equation either by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 1, b = -1, and c = 10.
x = (-(-1) ± √((-1)^2 - 4(1)(10))) / 2(1)
x = (1 ± √(1 - 40)) / 2
x = (1 ± √(-39)) / 2
Since the discriminant (√(b^2 - 4ac)) is negative, the equation has no real solutions. Therefore, there is no solution to the quadratic equation -x^2 + x = -10.
Which of the following coordinate pairs is a point on the parabola y=−4x^2−53x−56 ?
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To find the points on the parabola y = -4x^2 - 53x - 56, we can substitute different values of x into the equation to find the corresponding y-values. Let's evaluate the options using this method:
1) (0, -56)
When x = 0, y = -4(0)^2 - 53(0) - 56
y = 0 - 0 - 56
y = -56
2) (-2, -32)
When x = -2, y = -4(-2)^2 - 53(-2) - 56
y = -4(4) + 106 - 56
y = -16 + 106 - 56
y = 34
3) (1, -113)
When x = 1, y = -4(1)^2 - 53(1) - 56
y = -4 - 53 - 56
y = -113
4) (2, -180)
When x = 2, y = -4(2)^2 - 53(2) - 56
y = -4(4) - 106 - 56
y = -16 - 106 - 56
y = -180
Out of the given options, the coordinate pair (0, -56) is a point on the parabola y = -4x^2 - 53x - 56.
Which of the following coordinate pairs is a point on the parabola y=−4x2−53x−56 ?
(−4,−56) left parenthesis negative 4 comma negative 56 right parenthesis (−1,13) left parenthesis neagtive 1 comma 13 right parenthesis (−4,220) left parenthesis negative 4 comma 220 right parenthesis (−1,−7)
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To determine which of the given coordinate pairs is a point on the parabola y = -4x^2 - 53x - 56, we need to substitute the x-value into the equation and check if the resulting y-value matches the given y-value. Let's evaluate each option:
1) (-4, -56)
When x = -4, y = -4(-4)^2 - 53(-4) - 56
y = -4(16) + 212 - 56
y = -64 + 212 - 56
y = 92
The calculated y-value is 92, which does not match the given y-value of -56. Therefore, (-4, -56) is not a point on the parabola.
2) (-1, 13)
When x = -1, y = -4(-1)^2 - 53(-1) - 56
y = -4(1) + 53 - 56
y = -4 + 53 - 56
y = -7
The calculated y-value is -7, which matches the given y-value of 13. Therefore, (-1, 13) is a point on the parabola.
3) (-4, 220)
When x = -4, y = -4(-4)^2 - 53(-4) - 56
y = -4(16) + 212 - 56
y = -64 + 212 - 56
y = 92
The calculated y-value is 92, which does not match the given y-value of 220. Therefore, (-4, 220) is not a point on the parabola.
4) (-1, -7)
When x = -1, y = -4(-1)^2 - 53(-1) - 56
y = -4(1) + 53 - 56
y = -4 + 53 - 56
y = -7
The calculated y-value is -7, which matches the given y-value of -7. Therefore, (-1, -7) is a point on the parabola.
In conclusion, the coordinate pairs that are points on the parabola y = -4x^2 - 53x - 56 are (-1, 13) and (-1, -7).
To solve the quadratic equation x^2 + x + 0.25 = 0 using the perfect square trinomial pattern, we need to first determine if the equation can be factored. The perfect square trinomial pattern is applicable when the quadratic equation is in the form ax^2 + bx + c = 0, where a, b, and c are constants, and the coefficient of x^2 is 1.
In this case, we have x^2 + x + 0.25 = 0, and the coefficient of x^2 is already 1. Now, we can compare this equation to the perfect square trinomial pattern, which is (x + m)^2 = x^2 + 2mx + m^2.
By comparing the coefficients of the given equation with the pattern, we can see that 2mx = x and m^2 = 0.25. Since 2mx = x, m = 1/2.
Now, we can rewrite the equation x^2 + x + 0.25 as (x + 1/2)^2 = 0.25.
Taking the square root of both sides, we have x + 1/2 = ±√(0.25).
Simplifying, we get x + 1/2 = ±0.5.
To find the values of x, we can solve for two cases:
Case 1:
x + 1/2 = 0.5
Subtracting 1/2 from both sides, we have x = 0.5 - 1/2.
Simplifying, x = 0.
Case 2:
x + 1/2 = -0.5
Subtracting 1/2 from both sides, we have x = -0.5 - 1/2.
Simplifying, x = -1.
Therefore, the solutions to the quadratic equation x^2 + x + 0.25 = 0 are x = 0 and x = -1.