1.1.4 $$ 2 x-\sqrt{32-8 x}=0 $$ $$ 1.1 .5 \quad 2^{x}-5.2^{x+1}=-144 $$ 1.2 Solve simultaneously for $$ x $$ and $$ y $$ $$ \begin{array}{l} x-4 y=5 \text { and } \ x^{2}-2 x y-3 y^{2}=0 \end{array} $$ 1.3 The roots of a quadratic equation are given by: $$ x=\frac{-6 \pm \sqrt{(m+3)(5-m)}}{3} $$ Determine the values(s) of $$ m $$ for which the roots are non-re QUESTION 2 2.1 Simplify the following expression without a calculator

Question

UpStudy AI · Accepted Answer

Let's solve each part of your problem step by step.

### 1.1.4 Solve $$ 2x - \sqrt{32 - 8x} = 0 $$

1. Rearranging the equation gives:
   $$
   \sqrt{32 - 8x} = 2x
   $$
2. Squaring both sides results in:
   $$
   32 - 8x = (2x)^2
   $$
3. This simplifies to:
   $$
   32 - 8x = 4x^2
   $$
4. Rearranging gives:
   $$
   4x^2 + 8x - 32 = 0
   $$
5. Dividing the entire equation by 4 simplifies it to:
   $$
   x^2 + 2x - 8 = 0
   $$

Now, let's solve this quadratic equation.

### 1.1.5 Solve $$ 2^x - 5 \cdot 2^{x+1} = -144 $$

1. Rewrite the equation:
   $$
   2^x - 5 \cdot 2 \cdot 2^x = -144
   $$
   This simplifies to:
   $$
   2^x - 10 \cdot 2^x = -144
   $$
   or
   $$
   -9 \cdot 2^x = -144
   $$
2. Dividing both sides by -9 gives:
   $$
   2^x = 16
   $$
3. Taking the logarithm base 2:
   $$
   x = 4
   $$

### 1.2 Solve simultaneously for $$ x $$ and $$ y $$

The equations are:
$$
\begin{array}{l}
x - 4y = 5 \
x^2 - 2xy - 3y^2 = 0
\end{array}
$$

1. From the first equation, express $$ x $$ in terms of $$ y $$:
   $$
   x = 4y + 5
   $$
2. Substitute $$ x $$ into the second equation:
   $$
   (4y + 5)^2 - 2(4y + 5)y - 3y^2 = 0
   $$

Now, let's solve this system of equations.

### 1.3 Determine the values of $$ m $$ for which the roots are non-real

The roots are given by:
$$
x = \frac{-6 \pm \sqrt{(m+3)(5-m)}}{3}
$$

For the roots to be non-real, the discriminant must be less than zero:
$$
(m + 3)(5 - m) < 0
$$

Now, let's compute the solutions for each part.

I'll proceed with the calculations for parts 1.1.4, 1.1.5, 1.2, and 1.3 simultaneously.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$2^{x}=16$$
- step1: Rewrite in exponential form:
  $$2^{x}=2^{4}$$
- step2: Set the exponents equal:
  $$x=4$$
Solve the equation $$ (m + 3)(5 - m) < 0 $$.
Solve the inequality by following steps:
- step0: Solve the inequality by testing the values in the interval:
  $$\left(m+3\right)\left(5-m\right)\lt 0$$
- step1: Rewrite the expression:
  $$\left(m+3\right)\left(5-m\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&m+3=0\&5-m=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&m=-3\&m=5\end{align}$$
- step4: Determine the test intervals:
  $$\begin{align}&m<-3\&-35\end{align}$$
- step5: Choose a value:
  $$\begin{align}&m_{1}=-4\&m_{2}=1\&m_{3}=6\end{align}$$
- step6: Test the chosen value:
  $$\begin{align}&m<-3\textrm{ }\textrm{is the solution}\&-35\textrm{ }\textrm{is the solution}\end{align}$$
- step7: The final solution is $$m \in \left(-\infty,-3\right)\cup \left(5,+\infty\right):$$
  $$m \in \left(-\infty,-3\right)\cup \left(5,+\infty\right)$$
Solve the equation $$ x^2 + 2x - 8 = 0 $$.
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
  $$x^{2}+2x-8=0$$
- step1: Factor the expression:
  $$\left(x-2\right)\left(x+4\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&x-2=0\&x+4=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&x=2\&x=-4\end{align}$$
- step4: Rewrite:
  $$x_{1}=-4,x_{2}=2$$
Solve the system of equations $$ x - 4y = 5; x^2 - 2xy - 3y^2 = 0 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}x-4y=5\x^{2}-2xy-3y^{2}=0\end{array}\right.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=5+4y\x^{2}-2xy-3y^{2}=0\end{array}\right.$$
- step2: Substitute the value of $$x:$$
  $$\left(5+4y\right)^{2}-2\left(5+4y\right)y-3y^{2}=0$$
- step3: Simplify:
  $$25+30y+5y^{2}=0$$
- step4: Factor the expression:
  $$5\left(1+y\right)\left(5+y\right)=0$$
- step5: Divide the terms:
  $$\left(1+y\right)\left(5+y\right)=0$$
- step6: Separate into possible cases:
  $$\begin{align}&1+y=0\&5+y=0\end{align}$$
- step7: Solve the equation:
  $$\begin{align}&y=-1\&y=-5\end{align}$$
- step8: Calculate:
  $$y=-1\cup y=-5$$
- step9: Rearrange the terms:
  $$\left\{ \begin{array}{l}x=5+4y\y=-1\end{array}\right.\cup \left\{ \begin{array}{l}x=5+4y\y=-5\end{array}\right.$$
- step10: Calculate:
  $$\left\{ \begin{array}{l}x=1\y=-1\end{array}\right.\cup \left\{ \begin{array}{l}x=-15\y=-5\end{array}\right.$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}x=-15\y=-5\end{array}\right.\cup \left\{ \begin{array}{l}x=1\y=-1\end{array}\right.$$
- step12: Check the solution:
  $$\left\{ \begin{array}{l}x=-15\y=-5\end{array}\right.\cup \left\{ \begin{array}{l}x=1\y=-1\end{array}\right.$$
- step13: Rewrite:
  $$\left(x,y\right) = \left(-15,-5\right)\cup \left(x,y\right) = \left(1,-1\right)$$
Here are the solutions for each part of your problem:

### 1.1.4 Solve $$ 2x - \sqrt{32 - 8x} = 0 $$

The solutions to the equation $$ x^2 + 2x - 8 = 0 $$ are:
- $$ x_1 = -4 $$
- $$ x_2 = 2 $$

### 1.1.5 Solve $$ 2^x - 5 \cdot 2^{x+1} = -144 $$

The solution to the equation is:
- $$ x = 4 $$

### 1.2 Solve simultaneously for $$ x $$ and $$ y $$

The solutions to the system of equations are:
- $$ (x, y) = (-15, -5) $$
- $$ (x, y) = (1, -1) $$

### 1.3 Determine the values of $$ m $$ for which the roots are non-real

The values of $$ m $$ for which the roots are non-real are:
- $$ m \in (-\infty, -3) \cup (5, +\infty) $$

If you have any further questions or need additional assistance, feel free to ask!
Solutions: 1. **1.1.4** $$ x = -4 $$ or $$ x = 2 $$ 2. **1.1.5** $$ x = 4 $$ 3. **1.2** $$ (x, y) = (-15, -5) $$ or $$ (x, y) = (1, -1) $$ 4. **1.3** $$ m < -3 $$ or $$ m > 5 $$

1.1.4

1.2 Solve simultaneously for and

1.3 The roots of a quadratic equation are given by:

Determine the values(s) of for which the roots are non-re

QUESTION 2
2.1 Simplify the following expression without a calculator

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Solución

1.1.4 Solve

1.1.5 Solve

1.2 Solve simultaneously for and

1.3 Determine the values of for which the roots are non-real

1.1.4 Solve

1.1.5 Solve

1.2 Solve simultaneously for and

1.3 Determine the values of for which the roots are non-real

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Latest Algebra Questions

1.1.4 1.2 Solve simultaneously for and 1.3 The roots of a quadratic equation are given by: Determine the values(s) of for which the roots are non-re QUESTION 2 2.1 Simplify the following expression without a calculator

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Solución

1.1.4 Solve

1.1.5 Solve

1.2 Solve simultaneously for and

1.3 Determine the values of for which the roots are non-real

1.1.4 Solve

1.1.5 Solve

1.2 Solve simultaneously for and

1.3 Determine the values of for which the roots are non-real

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preguntas relacionadas

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1.1.4

1.2 Solve simultaneously for and

1.3 The roots of a quadratic equation are given by:

Determine the values(s) of for which the roots are non-re

QUESTION 2
2.1 Simplify the following expression without a calculator