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3 years ago

How do I solve this equation?

Mathematics

2 answers:

Ymorist [56]3 years ago

5 0

Answer:

x = 22/3, x=14

Step-by-step explanation:

Start by isolating whatever is in the absolute value sign; in this case divide either side by 4

4|1-3/4x + 7| =10,

----------------- ----

4 4

|1-3/4x + 7| = 10/4 = 5/2,

now if we apply the absolute value rule we know that 1-3/4x + 7 will be equal to -5/2, and 5/2

1-3/4x + 7 = 5/2 and 1-3/4x + 7 = -5/2

-3/4x + 8 = 5/2,

-3/4x=5/2-8,

-3/4x = 5/2-16/2,

-3/4x = -11/2,

x = -11/2*-4/3 = 44/6 = 22/3

-3/4x+8 = -5/2

-3/4x = -5/2-8

-3/4x = -5/2-16/2

-3/4x = -21/2

x = -21/2*-4/3 = 82/6 = 14

Veronika [31]3 years ago

4 0

Answer:

x=14

Step-by-step explanation:

Hope This Helps!

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Show work please<br> \sqrt(x+12)-\sqrt(2x+1)=1

Nesterboy [21]

Answer:

$x=4$

Step-by-step explanation:

Given $\displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1$ , start by squaring both sides to work towards isolating $x$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2$

Recall $(a-b)^2=a^2-2ab+b^2$ and $\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1$

Isolate the radical:

$\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}$

Square both sides:

$\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2$

Expand using FOIL and $(a+b)^2=a^2+2ab+b^2$ :

$\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36$

Move everything to one side to get a quadratic:

$\displaystyle-\frac{1}{4}x^2+7x-24=0$

Solving using the quadratic formula:

A quadratic in $ax^2+bx+c$ has real solutions $\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . In $\displaystyle-\frac{1}{4}x^2+7x-24$ , assign values:

$\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24$

Solving yields:

$\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}$

Only $x=4$ works when plugged in the original equation. Therefore, $x=24$ is extraneous and the only solution is $\boxed{x=4}$

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