Solve this quadratic equation. Return the answer in 2 decimals ?

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Solve this quadratic equation. Return the answer in 2 decimals ?

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Answer 1 · 2018-02-16T11:48:59

$x = 3.64, - 0.14$ $x=3.64,-0.14$

Explanation:

We have $2 x - \frac{1}{x} = 7$ $2x-1/x=7$

Multiplying both sides by $x$ $x$ , we get:

$x (2 x - \frac{1}{x}) = 7 x$ $x(2x-1/x)=7x$

$2 x^{2} - 1 = 7 x$ $2x^2-1=7x$

$2 x^{2} - 7 x - 1 = 0$ $2x^2-7x-1=0$

Now we have a quadratic equation. For any $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , where $a \neq 0,$ $a!=0,$ $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$ .

Here, $a = 2, b = - 7, c = - 1$ $a=2,b=-7,c=-1$

We can input:

$\frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 \cdot 2 \cdot - 1}}{2 \cdot 2}$ $(-(-7)+-sqrt((-7)^2-4*2*-1))/(2*2)$

$\frac{7 \pm \sqrt{49 + 8}}{4}$ $(7+-sqrt(49+8))/4$

$\frac{7 \pm \sqrt{57}}{4}$ $(7+-sqrt(57))/4$

$x = \frac{7 + \sqrt{57}}{4}, \frac{7 - \sqrt{57}}{4}$ $x=(7+sqrt(57))/4,(7-sqrt(57))/4$

$x = 3.64, - 0.14$ $x=3.64,-0.14$

Answer 2 · 2018-02-16T11:52:22

$x = 3.64 or x = - 0.14$ $x = 3.64 or x = -0.14$

Explanation:

This is clearly not a comfortable form to work with.
Multiply through by $x$ $x$ and re-arrange the equation into the form:

$a x^{2} + b x + c = 0$ $ax^2 +bx+c=0$

$2 x \times x - \frac{1}{x} \times x = 7 \times x$ $2xcolor(blue)(xx x) -1/xcolor(blue)(xx x) =7color(blue)(xx x)$

$2 x^{2} - 1 = 7 x$ $2x^2 -1=7x$

$2 x^{2} - 7 x - 1 = 0 \leftarrow$ $2x^2 -7x-1=0" "larr$ it does not factorise

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x= (-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 (2) (- 1)}}{2 (2)}$ $x = (-(-7)+-sqrt((-7)^2 -4(2)(-1)))/(2(2))$

$x = \frac{7 \pm \sqrt{49 + 8}}{4}$ $x = (7+-sqrt(49+8))/(4)$

$x = \frac{7 + \sqrt{57}}{4} = 3.64$ $x = (7+sqrt57)/4 = 3.64$

$x = \frac{7 - \sqrt{57}}{4} = - 0.14$ $x = (7-sqrt57)/4 = -0.14$

Answer 3 · 2018-02-16T11:53:47

See below...

Explanation:

First we need the standard format of $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

First we multiply all by $x$ $x$ to remove the fraction.

$2 x - \frac{1}{x} = 7 \Rightarrow 2 x^{2} - 1 = 7 x$ $2x-1/x=7 => 2x^2-1=7x$

Now we move the $7 x$ $7x$ over by subtracting both sides by $7 x$ $7x$

$2 x^{2} - 1 = 7 x \Rightarrow 2 x^{2} - 7 x - 1 = 0$ $2x^2-1=7x => 2x^2-7x-1=0$

As we want the answers to $2 d . p$ $2d.p$ it strongly hints that we need to use the quadratic formula.

We know that $x = - b \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a}$ $x=-b+-sqrt(b^2-4ac)/(2a)$

Now from our equation we know that ...

$a = 2$ $a =2$ , $b = - 7$ $b=-7$ and $c = - 1$ $c=-1$

Now we plug these into our formula, but as we have a $+$ $+$ and a $-$ $-$ we have to do it twice.

$x = - (- 7) + \frac{\sqrt{{(- 7)}^{2} - 4 (2) (- 1)}}{2 (2)}$ $x=-(-7)+sqrt((-7)^2-4(2)(-1))/(2(2))$
$x = - (- 7) - \frac{\sqrt{{(- 7)}^{2} - 4 (2) (- 1)}}{2 (2)}$ $x=-(-7)-sqrt((-7)^2-4(2)(-1))/(2(2))$

Now we put each one into our calculator and round to $2 d . p .$ $2d.p.$

$therefore x =-0.14 , x =3.64$

Both to $2d.p$

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