Combining Methods for Solving Quadratic Equations

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Question 1 of 5

1. Question
Solve for $x$ $x$

$3^{2 x} - 3^{x} - 72 = 0$ $3^(2x)-3^x-72=0$
- 1. $2$ $2$
- 2. $3$ $3$
- 3. $2$ $2$ and $4$ $4$
- 4. $2$ $2$ and $3$ $3$
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Reducible equations are non-quadratic equations that can be reduced into a quadratic equation for easier solving.

First, rewrite the equation as a quadratic equation by assigning a new variable

let

$u = 3^{x}$ $u=3^x$

$u^{2} = 3^{2 x}$ $u^2=3^(2x)$

$3^{2 x} - 3^{x} - 72$ $3^(2x)-3^x-72$ $=$ $=$ $0$ $0$

$u^{2} - u - 72$ $u^2-u-72$ $=$ $=$ $0$ $0$ Substitute new variable

Solve for $u$ $u$ by using cross method.

$(u - 9) (u + 8)$ $(u-9)(u+8)$ $=$ $=$ $0$ $0$

$u - 9$ $u-9$ $=$ $=$ $0$ $0$

$u - 9$ $u-9$ $+ 9$ $+9$ $=$ $=$ $0$ $0$ $+ 9$ $+9$

$u$ $u$ $=$ $=$ $9$ $9$

$u + 8$ $u+8$ $=$ $=$ $0$ $0$

$u + 8$ $u+8$ $- 8$ $-8$ $=$ $=$ $0$ $0$ $- 8$ $-8$

$u$ $u$ $=$ $=$ $- 8$ $-8$

Finally, substitute $u = 3^{x}$ $u=3^x$ to get the values of $x$ $x$

$u$ $u$ $=$ $=$ $9$ $9$

$3^{x}$ $3^x$ $=$ $=$ $9$ $9$ Substitute $u = 3^{x}$ $u=3^x$

$3^{x}$ $3^x$ $=$ $=$ $3^{2}$ $3^2$

$x$ $x$ $=$ $=$ $2$ $2$ Equal bases means equal exponents

$u$ $u$ $=$ $=$ $- 8$ $-8$

$3^{x}$ $3^x$ $=$ $=$ $- 8$ $-8$ Substitute $u = 3^{x}$ $u=3^x$

This has no solution since there is no $x$ $x$ value that can make $3^{x}$ $3^x$ negative

$x = 2$ $x=2$
Question 2 of 5

2. Question
Solve for $x$ $x$

$2^{2 x} - {3.2}^{x} - 40 = 0$ $2^(2x)-3.2^x-40=0$
- 1. $3$ $3$
- 2. $4$ $4$
- 3. $2$ $2$ and $3$ $3$
- 4. $2$ $2$ and $8$ $8$
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Reducible equations are non-quadratic equations that can be reduced into a quadratic equation for easier solving.

First, rewrite the equation as a quadratic equation by assigning a new variable

let

$u = 2^{x}$ $u=2^x$

$2^{2 x} - {3.2}^{x} - 40$ $2^(2x)-3.2^x-40$ $=$ $=$ $0$ $0$

$u^{2} - 3 u - 40$ $u^2-3u-40$ $=$ $=$ $0$ $0$ Substitute new variable

Solve for $u$ $u$ by using cross method.

$(u + 5) (u - 8)$ $(u+5)(u-8)$ $=$ $=$ $0$ $0$

$u + 5$ $u+5$ $=$ $=$ $0$ $0$

$u + 5$ $u+5$ $- 5$ $-5$ $=$ $=$ $0$ $0$ $- 5$ $-5$

$u$ $u$ $=$ $=$ $- 5$ $-5$

$u - 8$ $u-8$ $=$ $=$ $0$ $0$

$u - 8$ $u-8$ $+ 8$ $+8$ $=$ $=$ $0$ $0$ $+ 8$ $+8$

$u$ $u$ $=$ $=$ $8$ $8$

Finally, substitute $u = 2^{x}$ $u=2^x$ to get the values of $x$ $x$

$u$ $u$ $=$ $=$ $8$ $8$

$2^{x}$ $2^x$ $=$ $=$ $8$ $8$ Substitute $u = 2^{x}$ $u=2^x$

$2^{x}$ $2^x$ $=$ $=$ $2^{3}$ $2^3$

$x$ $x$ $=$ $=$ $3$ $3$ Equal bases means equal exponents

$u$ $u$ $=$ $=$ $- 5$ $-5$

$2^{x}$ $2^x$ $=$ $=$ $- 5$ $-5$ Substitute $u = 2^{x}$ $u=2^x$

This has no solution since there is no $x$ $x$ value that can make $2^{x}$ $2^x$ negative

$x = 3$ $x=3$
Question 3 of 5

3. Question
Solve for $x$ $x$

$x^{4} - 7 x^{2} - 18 = 0$ $x^4-7x^2-18=0$
- 1. $3$ $3$
- 2. $- 3$ $-3$
- 3. $3$ $3$ and $- 3$ $-3$
- 4. No Solution
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Reducible equations are non-quadratic equations that can be reduced into a quadratic equation for easier solving.

First, rewrite the equation as a quadratic equation by assigning a new variable

let

$u = x^{2}$ $u=x^2$

$x^{4} - 7 x^{2} - 18$ $x^4-7x^2-18$ $=$ $=$ $0$ $0$

$u^{2} - 7 u - 18$ $u^2-7u-18$ $=$ $=$ $0$ $0$ Substitute new variable

Solve for $u$ $u$ by using cross method.

$(u - 9) (u + 2)$ $(u-9)(u+2)$ $=$ $=$ $0$ $0$

$u - 9$ $u-9$ $=$ $=$ $0$ $0$

$u - 9$ $u-9$ $+ 9$ $+9$ $=$ $=$ $0$ $0$ $+ 9$ $+9$

$u$ $u$ $=$ $=$ $9$ $9$

$u + 2$ $u+2$ $=$ $=$ $0$ $0$

$u + 2$ $u+2$ $- 2$ $-2$ $=$ $=$ $0$ $0$ $- 2$ $-2$

$u$ $u$ $=$ $=$ $- 2$ $-2$

Finally, substitute $u = x^{2}$ $u=x^2$ to get the values of $x$ $x$

$u$ $u$ $=$ $=$ $9$ $9$

$x^{2}$ $x^2$ $=$ $=$ $9$ $9$ Substitute $u = x^{2}$ $u=x^2$

$\sqrt{x^{2}}$ $sqrt(x^2)$ $=$ $=$ $\sqrt{9}$ $sqrt9$ Get the square root of both sides

$x$ $x$ $=$ $=$ $\pm 3$ $+-3$

$u$ $u$ $=$ $=$ $- 2$ $-2$

$x^{2}$ $x^2$ $=$ $=$ $- 2$ $-2$ Substitute $u = x^{2}$ $u=x^2$

This has no solution since there is no $x$ $x$ value that can make $x^{2}$ $x^2$ negative

$x = 3, - 3$ $x=3, -3$
Question 4 of 5

4. Question
Solve for $x$ $x$

$4 x^{4} + 3 x^{2} - 10 = 0$ $4x^4+3x^2-10=0$
- 1. $\sqrt{5}$ $sqrt5$ and $- \sqrt{5}$ $-sqrt5$
- 2. $\frac{\sqrt{2}}{5}$ $(sqrt2)/5$ and $- \frac{\sqrt{2}}{5}$ $-(sqrt2)/5$
- 3. $\sqrt{2}$ $sqrt2$ and $- \sqrt{2}$ $-sqrt2$
- 4. $\frac{\sqrt{5}}{2}$ $(sqrt5)/2$ and $- \frac{\sqrt{5}}{2}$ $-(sqrt5)/2$
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Reducible equations are non-quadratic equations that can be reduced into a quadratic equation for easier solving.

First, rewrite the equation as a quadratic equation by assigning a new variable

let

$u = x^{2}$ $u=x^2$

$4 x^{4} + 3 x^{2} - 10$ $4x^4+3x^2-10$ $=$ $=$ $0$ $0$

$4 u^{2} + 3 u - 10$ $4u^2+3u-10$ $=$ $=$ $0$ $0$ Substitute new variable

Solve for $u$ $u$ by using cross method.

$(4 u - 5) (u + 2)$ $(4u-5)(u+2)$ $=$ $=$ $0$ $0$

$4 u - 5$ $4u-5$ $=$ $=$ $0$ $0$

$4 u - 5$ $4u-5$ $+ 5$ $+5$ $=$ $=$ $0$ $0$ $+ 5$ $+5$

$4 u$ $4u$ $=$ $=$ $5$ $5$

$u$ $u$ $=$ $=$ $\frac{5}{4}$ $5/4$

$u + 2$ $u+2$ $=$ $=$ $0$ $0$

$u + 2$ $u+2$ $- 2$ $-2$ $=$ $=$ $0$ $0$ $- 2$ $-2$

$u$ $u$ $=$ $=$ $- 2$ $-2$

Finally, substitute $u = x^{2}$ $u=x^2$ to get the values of $x$ $x$

$u$ $u$ $=$ $=$ $\frac{5}{4}$ $5/4$

$x^{2}$ $x^2$ $=$ $=$ $\frac{5}{4}$ $5/4$ Substitute $u = x^{2}$ $u=x^2$

$\sqrt{x^{2}}$ $sqrt(x^2)$ $=$ $=$ $\sqrt{\frac{5}{4}}$ $sqrt(5/4)$ Get the square root of both sides

$x$ $x$ $=$ $=$ $\pm \frac{\sqrt{5}}{2}$ $+-(sqrt5)/2$

$u$ $u$ $=$ $=$ $- 2$ $-2$

$x^{2}$ $x^2$ $=$ $=$ $- 2$ $-2$ Substitute $u = x^{2}$ $u=x^2$

This has no solution since there is no $x$ $x$ value that can make $x^{2}$ $x^2$ negative

$x = \frac{\sqrt{5}}{2}, - \frac{\sqrt{5}}{2}$ $x=(sqrt5)/2, -(sqrt5)/2$

Question 5 of 5

Solve for

x

$x$

4^{x} + 3 \cdot 2^{x} - 10 = 0

$4^x+3*2^x-10=0$

1. $1$ $1$
2. $5$ $5$
3. $- 5$ $-5$
4. $1$ $1$ and $5$ $5$

Combining Methods for Solving Quadratic Equations

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1. Question

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2. Question

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3. Question

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4. Question

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5. Question

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Quizzes

$3^{2 x} - 3^{x} - 72$ $3^(2x)-3^x-72$	$=$ $=$	$0$ $0$
$u^{2} - u - 72$ $u^2-u-72$	$=$ $=$	$0$ $0$	Substitute new variable

$u - 9$ $u-9$	$=$ $=$	$0$ $0$
$u - 9$ $u-9$ $+ 9$ $+9$	$=$ $=$	$0$ $0$ $+ 9$ $+9$
$u$ $u$	$=$ $=$	$9$ $9$

$u + 8$ $u+8$	$=$ $=$	$0$ $0$
$u + 8$ $u+8$ $- 8$ $-8$	$=$ $=$	$0$ $0$ $- 8$ $-8$
$u$ $u$	$=$ $=$	$- 8$ $-8$

$2^{2 x} - {3.2}^{x} - 40$ $2^(2x)-3.2^x-40$	$=$ $=$	$0$ $0$
$u^{2} - 3 u - 40$ $u^2-3u-40$	$=$ $=$	$0$ $0$	Substitute new variable

$u + 5$ $u+5$	$=$ $=$	$0$ $0$
$u + 5$ $u+5$ $- 5$ $-5$	$=$ $=$	$0$ $0$ $- 5$ $-5$
$u$ $u$	$=$ $=$	$- 5$ $-5$

$u - 8$ $u-8$	$=$ $=$	$0$ $0$
$u - 8$ $u-8$ $+ 8$ $+8$	$=$ $=$	$0$ $0$ $+ 8$ $+8$
$u$ $u$	$=$ $=$	$8$ $8$

$x^{4} - 7 x^{2} - 18$ $x^4-7x^2-18$	$=$ $=$	$0$ $0$
$u^{2} - 7 u - 18$ $u^2-7u-18$	$=$ $=$	$0$ $0$	Substitute new variable

$u + 2$ $u+2$	$=$ $=$	$0$ $0$
$u + 2$ $u+2$ $- 2$ $-2$	$=$ $=$	$0$ $0$ $- 2$ $-2$
$u$ $u$	$=$ $=$	$- 2$ $-2$

$4 x^{4} + 3 x^{2} - 10$ $4x^4+3x^2-10$	$=$ $=$	$0$ $0$
$4 u^{2} + 3 u - 10$ $4u^2+3u-10$	$=$ $=$	$0$ $0$	Substitute new variable

$4 u - 5$ $4u-5$	$=$ $=$	$0$ $0$
$4 u - 5$ $4u-5$ $+ 5$ $+5$	$=$ $=$	$0$ $0$ $+ 5$ $+5$
$4 u$ $4u$	$=$ $=$	$5$ $5$
$u$ $u$	$=$ $=$	$\frac{5}{4}$ $5/4$

$u$ $u$	$=$ $=$	$\frac{5}{4}$ $5/4$
$x^{2}$ $x^2$	$=$ $=$	$\frac{5}{4}$ $5/4$	Substitute $u = x^{2}$ $u=x^2$
$\sqrt{x^{2}}$ $sqrt(x^2)$	$=$ $=$	$\sqrt{\frac{5}{4}}$ $sqrt(5/4)$	Get the square root of both sides
$x$ $x$	$=$ $=$	$\pm \frac{\sqrt{5}}{2}$ $+-(sqrt5)/2$

$u$ $u$	$=$ $=$	$2$ $2$
$2^{x}$ $2^x$	$=$ $=$	$2$ $2$	Substitute $u = 2^{x}$ $u=2^x$
$2^{x}$ $2^x$	$=$ $=$	$2^{1}$ $2^1$
$x$ $x$	$=$ $=$	$1$ $1$	Equal bases means equal exponents

$4^{x} + 3 \cdot 2^{x} - 10$ $4^x+3*2^x-10$	$=$ $=$	$0$ $0$
${(2^{2})}^{x} + 3 \cdot 2^{x} - 10$ $(2^2)^x+3*2^x-10$	$=$ $=$	$0$ $0$
$2^{2 x} + 3 \cdot 2^{x} - 10$ $2^(2x)+3*2^x-10$	$=$ $=$	$0$ $0$
$u^{2} + 3 u - 10$ $u^2+3u-10$	$=$ $=$	$0$ $0$	Substitute new variable

$u - 2$ $u-2$	$=$ $=$	$0$ $0$
$u - 2$ $u-2$ $+ 2$ $+2$	$=$ $=$	$0$ $0$ $+ 2$ $+2$
$u$ $u$	$=$ $=$	$2$ $2$