Solving Algebraic Problems with Quadratic Equations
a) Show that the solutions of the equation \(x^2 - 6x - 43 = 0\) can be written in the form \(x = p \pm \frac{q\sqrt{13}}{13}\), where \(p\) and \(q\) are positive integers.
b) Based on the above or any other method, solve the inequality \(x^2 - 6x - 43 \le 0\).
A function \(f\) is defined by \(f(x) = 3(x-1)^2 - 18\), \(x \in \mathbb{R}\).
a) Write \(f(x)\) in the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants.
b) Find the coordinates of the vertex of the graph of \(f\).
c) Find the equation of the axis of symmetry of the graph of \(f\).
d) Indicate the range of \(f\).
e) The graph of \(f\) is translated by the vector \(\begin{pmatrix} 2 \\ -1 \end{pmatrix}\) to form a new curve that represents a new function \(g(x)\).
Find \(g(x)\) in the form \(px^2 + qx + r\), where \(p\), \(q\), and \(r\) are constants.
a) Solve the equation \(8x^2 + 6x - 5 = 0\) by factorization.
b) Determine the range of values of \(k\) for which the equation \(8x^2 + 6x - 5 = k\) has no real solutions.
Consider the function \(f(x) = -x^2 - 10x + 27\), \(x \in \mathbb{R}\).
a) Show that the function \(f\) can be expressed in the form \(f(x) = a(x-h)^2 + k\), where \(a\), \(h\), and \(k\) are constants.
b) Based on the above, write the coordinates of the vertex of the graph of \(y = f(x)\).
c) Based on the above, write the equation of the axis of symmetry of the graph of \(y = f(x)\).
The quadratic curve \(y = x^2 + bx + c\) cuts the x-axis at \( (10, 0) \) and has the equation of the line of symmetry \(x = \frac{5}{2}\).
a) Find the values of \(b\) and \(c\).
b) Based on the above, or any other method, find the other two coordinates where the curve intersects the y-axis.
Consider the function \(f(x) = 2x^2 - 4x - 8\), \(x \in \mathbb{R}\).
a) Show that the function \(f\) can be expressed in the form \(f(x) = a(x-h)^2 + k\), where \(a\), \(h\), and \(k\) are constants.
b) The function \(f(x)\) can be obtained through a sequence of transformations of \(g(x) = x^2\). Describe each transformation in order.
Consider the equation \(f(x) = 2kx^2 + 6x + k\), \(x \in \mathbb{R}\).
a) For the case where the equation \(f(x) = 0\) has two equal real roots, find the possible values of \(k\).
b) For the case where the equation of the axis of symmetry of the curve \(y = f(x)\) is \(x + 1 = 0\), find the value of \(k\).
c) Solve the equation \(f(x) = 0\) when \(k = 2\).
A curve \(y = f(x)\) passes through the points with coordinates \(A(-12, 10)\), \(B(0, -16)\), \(C(2, 9)\), and \(D(14, -10)\).
a) Write the coordinates of each point after the curve has been transformed by \(f(x) \rightarrow f(2x)\).
b) Write the coordinates of each point after the curve has been transformed by \(f(x) \rightarrow f(-x) + 3\).
Solving Algebraic Problems with Quadratic Equations
Problem 1:
a)
To show that the solutions of the equation (x^2 - 6x - 43 = 0) can be written in the form (x = p \pm \frac{q\sqrt{13}}{13}), we first find the roots using the quadratic formula:
[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-43)}}{2(1)} ]
[ x = \frac{6 \pm \sqrt{36 + 172}}{2} ]
[ x = \frac{6 \pm \sqrt{208}}{2} ]
[ x = \frac{6 \pm 4\sqrt{13}}{2} ]
[ x = 3 \pm 2\sqrt{13} ]
Hence, we can write ( x = p \pm \frac{q\sqrt{13}}{13} ) with ( p = 3 ) and ( q = 2 ).
b)
To solve the inequality (x^2 - 6x - 43 \le 0), we find the critical points by setting the expression equal to zero and solving for (x):
[ x^2 - 6x - 43 = 0 ]
[ (x - 9)(x + 3) = 0 ]
This gives us critical points at (x = 9) and (x = -3). We then test the intervals created by these points to determine where the inequality holds true. The solution is ( -3 \le x \le 9 ).
Problem 2:
a)
Given the function ( f(x) = 3(x-1)^2 - 18 ), we expand and simplify to express it in the form ( ax^2 + bx + c ):
[ f(x) = 3(x^2 - 2x + 1) - 18 ]
[ f(x) = 3x^2 - 6x + 3 - 18 ]
[ f(x) = 3x^2 - 6x - 15 ]
Therefore, ( a = 3, b = -6, c = -15 ).
b)
The vertex of a quadratic function in the form ( ax^2 + bx + c ) is given by ( (-\frac{b}{2a}, f(-\frac{b}{2a})) ). Substituting the values, we get the vertex as ( (1, -15) ).
c)
The equation of the axis of symmetry is given by ( x = -\frac{b}{2a} ). Substituting values, we have ( x = \frac{6}{6} = 1 ).
d)
The range of a quadratic function in the form ( ax^2 + bx + c ) where ( a > 0 ) is ( [f(h), +\infty) ), where ( f(h) ) is the minimum value of the function. In this case, since ( a = 3 > 0 ), the range is ( [-15, +\infty) ).
e)
To find the new function ( g(x) ) after translating by the vector ( (2, -1) ), we replace ( x ) with ( x-2 ) and ( y ) with ( y+1 ):
[ g(x) = f(x-2)+1 = 3((x-2)-1)^2 - 18 + 1 = 3(x-3)^2 - 17 ]
Therefore, ( g(x) = 3x^2 - 18x +19 ).
These steps provide solutions to the presented algebraic problems involving quadratic equations and functions.