Question

1. solve each equation by graphing the expressions on both sides of the

equation.
a) ( 3a^2 = 18a - 21 )
b) ( 5p = 3 - 2p^2 )
c) ( 4x(x + 3) = 3(4x + 3) )
d) ( x^2 - 3x - 8 = -2x^2 + 8x + 1 )

Explanation:

Part (a)

Step 1: Rearrange the equation

First, we rewrite the equation \(3a^{2}=18a - 21\) in standard quadratic form \(ax^{2}+bx + c = 0\). Subtract \(18a\) and add \(21\) to both sides:
\(3a^{2}-18a + 21=0\)
We can simplify this equation by dividing all terms by \(3\):
\(a^{2}-6a + 7 = 0\)

Step 2: Analyze the quadratic function

To solve by graphing, we consider the function \(y=a^{2}-6a + 7\). The graph of a quadratic function \(y = Ax^{2}+Bx + C\) is a parabola. For \(y=a^{2}-6a + 7\), \(A = 1\) (positive, so the parabola opens upwards), the vertex of the parabola is at \(a=-\frac{B}{2A}=-\frac{-6}{2\times1} = 3\). Substitute \(a = 3\) into the function: \(y=(3)^{2}-6\times3 + 7=9 - 18 + 7=-2\). So the vertex is at \((3,-2)\).

We can also find the \(y\)-intercept by setting \(a = 0\): \(y=0^{2}-6\times0 + 7 = 7\), so the \(y\)-intercept is \((0,7)\).

To find the \(x\)-intercepts (solutions of the equation \(a^{2}-6a + 7 = 0\)), we can use the quadratic formula \(a=\frac{-B\pm\sqrt{B^{2}-4AC}}{2A}\). Here, \(A = 1\), \(B=-6\), \(C = 7\). Then \(B^{2}-4AC=(-6)^{2}-4\times1\times7=36 - 28 = 8\). So \(a=\frac{6\pm\sqrt{8}}{2}=\frac{6\pm2\sqrt{2}}{2}=3\pm\sqrt{2}\approx3\pm1.414\), so \(a\approx4.414\) or \(a\approx1.586\).

When we graph \(y=a^{2}-6a + 7\), the \(x\)-intercepts (where \(y = 0\)) are the solutions to the original equation. So the solutions are \(a=3+\sqrt{2}\) and \(a=3-\sqrt{2}\) (or approximately \(a\approx4.41\) and \(a\approx1.59\)).

Part (b)

Step 1: Rearrange the equation

Rewrite the equation \(5p=3 - 2p^{2}\) in standard quadratic form. Add \(2p^{2}\) and subtract \(3\) from both sides:
\(2p^{2}+5p - 3=0\)

Step 2: Analyze the quadratic function

Consider the function \(y = 2p^{2}+5p - 3\). The parabola opens upwards since \(A = 2>0\). The vertex is at \(p=-\frac{B}{2A}=-\frac{5}{2\times2}=-\frac{5}{4}=-1.25\). Substitute \(p = - 1.25\) into the function: \(y=2\times(-1.25)^{2}+5\times(-1.25)-3=2\times1.5625-6.25 - 3=3.125-6.25 - 3=-6.125\). So the vertex is at \((-1.25,-6.125)\).

The \(y\)-intercept is at \(p = 0\): \(y=2\times0^{2}+5\times0 - 3=-3\), so \((0,-3)\).

Using the quadratic formula for \(2p^{2}+5p - 3=0\) where \(A = 2\), \(B = 5\), \(C=-3\). \(B^{2}-4AC=25-4\times2\times(-3)=25 + 24 = 49\). Then \(p=\frac{-5\pm\sqrt{49}}{2\times2}=\frac{-5\pm7}{4}\). We have two solutions: \(p=\frac{-5 + 7}{4}=\frac{2}{4}=0.5\) and \(p=\frac{-5 - 7}{4}=\frac{-12}{4}=-3\).

When we graph \(y = 2p^{2}+5p - 3\), the \(x\)-intercepts (where \(y = 0\)) are at \(p=-3\) and \(p = 0.5\), which are the solutions to the original equation.

Part (c)

Step 1: Expand and rearrange the equation

First, expand both sides of \(4x(x + 3)=3(4x + 3)\).

Left side: \(4x(x + 3)=4x^{2}+12x\)

Right side: \(3(4x + 3)=12x+9\)

Now, set the equation: \(4x^{2}+12x=12x + 9\)

Subtract \(12x\) and \(9\) from both sides to get standard quadratic form: \(4x^{2}-9=0\)

This can be factored as a difference of squares: \((2x - 3)(2x + 3)=0\)

Step 2: Analyze the function for graphing

To solve by graphing, consider the function \(y = 4x^{2}-9\). This is a quadratic function with \(A = 4\) (opens upwards), \(B = 0\), \(C=-9\). The vertex is at \(x =-\frac{B}{2A}=0\), and \(y=4\times0^{2}-9=-9\), so the vertex is at \((0,-9)\).

The \(y\)-intercept is at \(x = 0\), \(y=-9\) (which is the vertex in this case since \(B = 0\)).

To find the \(x\)-intercepts, set \(y = 0\): \(4x^{2}-9=0\Rightarrow4x^{2}=9\Rightarrow x^{2}=\frac{9}{4}\Rightarrow x=\pm\frac{3}{2}\). So the \(x\)-intercepts are at \(x=\frac{3}{2}=1.5\) and \(x =-\frac{3}{2}=-1.5\).

When we graph \(y = 4x^{2}-9\), the \(x\)-intercepts (where \(y = 0\)) are at \(x=-1.5\) and \(x = 1.5\), which are the solutions to the original equation.

Part (d)

Answer:

Step 1: Rearrange the equation

Start with \(x^{2}-3x - 8=-2x^{2}+8x + 1\)

Add \(2x^{2}\), subtract \(8x\), and subtract \(1\) from both sides to get standard quadratic form:

\(x^{2}+2x^{2}-3x-8x - 8 - 1=0\)

\(3x^{2}-11x - 9=0\)

Step 2: Analyze the quadratic function

To solve by graphing, consider the function \(y = 3x^{2}-11x - 9\). For a quadratic function \(y=Ax^{2}+Bx + C\), \(A = 3\) (opens upwards), the vertex is at \(x=-\frac{B}{2A}=-\frac{-11}{2\times3}=\frac{11}{6}\approx1.833\). Substitute \(x=\frac{11}{6}\) into the function:

\(y=3\times(\frac{11}{6})^{2}-11\times\frac{11}{6}-9=3\times\frac{121}{36}-\frac{121}{6}-9=\frac{121}{12}-\frac{242}{12}-\frac{108}{12}=\frac{121 - 242 - 108}{12}=\frac{-229}{12}\approx - 19.083\)

The \(y\)-intercept is at \(x = 0\): \(y=3\times0^{2}-11\times0 - 9=-9\), so \((0,-9)\).

Using the quadratic formula for \(3x^{2}-11x - 9=0\) where \(A = 3\), \(B=-11\), \(C=-9\). \(B^{2}-4AC=(-11)^{2}-4\times3\times(-9)=121 + 108 = 229\). Then \(x=\frac{11\pm\sqrt{229}}{2\times3}=\frac{11\pm\sqrt{229}}{6}\approx\frac{11\pm15.1327}{6}\). So \(x\approx\frac{11 + 15.1327}{6}\approx\frac{26.1327}{6}\approx4.355\) and \(x\approx\frac{11-15.1327}{6}\approx\frac{-4.1327}{6}\approx - 0.689\).

When we graph \(y = 3x^{2}-11x - 9\), the \(x\)-intercepts (where \(y = 0\)) are the solutions to the original equation, approximately \(x\approx4.36\) and \(x\approx - 0.69\).

Final Answers:

a) \(a = 3\pm\sqrt{2}\) (or approximately \(a\approx1.59\) and \(a\approx4.41\))
b) \(p=-3\) or \(p = 0.5\)
c) \(x=\pm\frac{3}{2}\) (or \(x = - 1.5\) and \(x = 1.5\))
d) \(x=\frac{11\pm\sqrt{229}}{6}\) (or approximately \(x\approx - 0.69\) and \(x\approx4.36\))