Below are the answers to each of your questions:
Q5 Determine whether 34 is the solution to the following quadratic equation \( 8x^2 - 6x - 3 = 0 \).
To determine if \( x = 34 \) is a solution, we will substitute \( x \) in the equation and check if the left-hand side equals zero.
\[ 8(34)^2 - 6(34) - 3 = 8(1156) - 204 - 3 \] \[ = 9248 - 204 - 3 \] \[ = 9241 \]
Since \( 9241 \neq 0 \), 34 is not a solution to the equation.
Q6 Determine whether -1 is a solution to the following quadratic equation \( 4x^2 + 4x + 1 = 0 \).
We will substitute \( x = -1 \) into the equation:
\[ 4(-1)^2 + 4(-1) + 1 = 4(1) - 4 + 1 \] \[ = 4 - 4 + 1 \] \[ = 1 \]
Since \( 1 \neq 0 \), -1 is not a solution to the equation.
Q7 Identify the vertex of \( y = -5(x - 8)^2 + 12 \).
This equation is already in vertex form \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex.
Here, \( h = 8 \) and \( k = 12 \).
Thus, the vertex is \( (8, 12) \).
Q8 Write the following equation in vertex form \( y = x^2 + 6x - 1 \).
To convert the equation to vertex form, we complete the square.
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Start with the equation: \[ y = x^2 + 6x - 1 \]
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Focus on the quadratic and linear terms: \[ y = (x^2 + 6x) - 1 \]
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Complete the square:
- Take half of the coefficient of \( x \) (which is 6), square it: \( (6/2)^2 = 9 \).
- Add and subtract this square inside the parentheses: \[ y = (x^2 + 6x + 9 - 9) - 1 \] \[ = (x + 3)^2 - 10 \]
Thus, the vertex form of the equation is: \[ y = (x + 3)^2 - 10 \]
If you have any more questions or need further clarification, feel free to ask!
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