How do you solve using the completing the square method $5 x^{2} + 3 x - 1 = 0$ $5x^2 + 3x - 1 = 0$ ?

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How do you solve using the completing the square method $5 x^{2} + 3 x - 1 = 0$ $5x^2 + 3x - 1 = 0$ ?

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Answer 1 · 2016-02-28T16:56:03

Given expression is
$5 x^{2} + 3 x - 1 = 0$ $5x^2+3x−1=0$

We are required to solve for $x$ $x$ by using *Completing the Square method.

Observe that first term can be made a perfect square if it is multiplied by $5$ $5$ . Therefore multiply $⋄$ $diamond$ both sides by $5$ $5$ .

$5 \times (5 x^{2} + 3 x - 1) = 5 \times 0$ $5xx(5x^2+3x−1)=5xx0$
$\Rightarrow 25 x^{2} + 15 x - 5 = 0$ $=> 25x^2+15x−5=0$
$\Rightarrow {(5 x)}^{2} + 15 x - 5 = 0$ $=> (5x)^2+15x−5=0$

Remember that ${(a x + b)}^{2} = {(a x)}^{2} + 2 a b x + b^{2}$ $(ax+b)^2=(ax)^2+2abx+b^2$ .
We need to make the second term $15 x$ $15x$ look like $2 a b x$ $2abx$ .
While comparing the coefficient of $x^{2}$ $x^2$ we see that $a = 5$ $a=5$ .
Therefore, $15 x = 2 a b x = 2 \cdot 5 \cdot \frac{3}{2} x$ $15x=2abx=2cdot5cdot3/2x$ . With this substitution our equation becomes
${(5 x)}^{2} + 2 \cdot 5 \cdot \frac{3}{2} x - 5 = 0$ $(5x)^2+2cdot5cdot3/2x−5=0$
Now to make a perfect square we need to make the constant term as $b^{2}$ $b^2$ .
We see that $b = \frac{3}{2}$ $b=3/2$ . Easiest way to do this is take the third term $- 5$ $-5$ in the problem to the other side With this our equation becomes
${(5 x)}^{2} + 2 \cdot 5 \cdot \frac{3}{2} x = 5$ $(5x)^2+2cdot5cdot3/2x=5$
and then add $b^{2} = {(\frac{3}{2})}^{2}$ $b^2=(3/2)^2$ to both sides. With this our equation becomes

${(5 x)}^{2} + 2 \cdot 5 \cdot \frac{3}{2} x + {(\frac{3}{2})}^{2} = 5 + {(\frac{3}{2})}^{2}$ $(5x)^2+2cdot5cdot3/2x+(3/2)^2=5+(3/2)^2$
Simplify right side of the equation and putting left hand side as perfect square
${(5 x)}^{2} + 2 \cdot 5 \cdot \frac{3}{2} x + {(\frac{3}{2})}^{2} = 5 + \frac{9}{4}$ $(5x)^2+2cdot5cdot3/2x+(3/2)^2=5+9/4$
${(5 x + \frac{3}{2})}^{2} = \frac{29}{4}$ $(5x+3/2)^2=29/4$
Taking square root of both sides
$5 x + \frac{3}{2} = \pm \sqrt{\frac{29}{4}}$ $5x+3/2=+-sqrt(29/4)$
Solving for $x$ $x$
$x = \frac{- \frac{3}{2} \pm \sqrt{\frac{29}{4}}}{5}$ $x=(-3/2+-sqrt(29/4))/5$ Simplifying
$x = \frac{- 3 \pm \sqrt{29}}{10}$ $x=(-3+-sqrt29)/10$
-.-.-.-.-.-.-.-.-.-.-.-.-..-.-.-.-.-.-..--

$⋄$ $diamond$ Note that the first term can be made a perfect square if it is divided by $5$ $5$ also.

You may divide both sides of equation with $5$ $5$ to obtain the solution. In such a case involvement of fractions in the solution will be more compared to multiplication option

Verify that we got the correct answer by inserting in

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

Answer 2 · 2016-02-29T02:24:01

$x = - \frac{3}{10} \pm \frac{\sqrt{29}}{10}$ $x = -3/10 +- sqrt29/10$

Explanation:

$x^{2} + \frac{3}{5} x = \frac{1}{5}$ $x^2 + 3/5 x = 1/5$
$(x^{2} + \frac{3}{5} x) + \frac{9}{100} = \frac{1}{5} + \frac{9}{100} = \frac{29}{100}$ $(x^2 + 3/5x) + 9/100 = 1/5 + 9/100 = 29/100$
${(x + \frac{3}{10})}^{2} = \frac{29}{100}$ $(x + 3/10)^2 = 29/100$
$(x + \frac{3}{10}) = \pm \frac{\sqrt{29}}{10}$ $(x + 3/10) = +- sqrt29/10$
$x = - \frac{3}{10} \pm \frac{\sqrt{29}}{10}$ $x = -3/10 +- sqrt29/10$

Answer 3 · 2016-03-06T15:05:57

Multiply the quadratic equation through by $20$ $20$ first to make the arithmetic work out better and find:

$x = \frac{- 3 \pm \sqrt{29}}{10}$ $x = (-3+-sqrt(29))/10$

Explanation:

Below, we use the difference of square identity:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

with $a = 10 x + 3$ $a=10x+3$ and $b = \sqrt{29}$ $b=sqrt(29)$

$color(white)()$
When asked to solve $5 x^{2} + 3 x - 1 = 0$ $5x^2+3x-1 = 0$ by "completing the square", there are two red flags:

The leading coefficient is not a perfect square.

The middle coefficient is not an even number.

This will lead to working with fractions and multiplying or dividing the whole equation by some factor. Either that or working with irrational coefficients, e.g. $5 x^{2} = {(\sqrt{5} x)}^{2}$ $5x^2 = (sqrt(5)x)^2$ .

To make things work out nicer, multiply the whole equation up front by $20 = 2^{2} \cdot 5$ $20 = 2^2 * 5$ to get:

$100 x^{2} + 60 x - 20 = 0$ $100x^2+60x-20 = 0$

Now the leading term is a square ( ${(10 x)}^{2} = 100 x^{2}$ $(10x)^2 = 100x^2$ ), we find:

$0 = 100 x^{2} + 60 x - 20$ $0 = 100x^2+60x-20$

$= {(10 x + 3)}^{2} - 9 - 20$ $=(10x+3)^2-9-20$

$= {(10 x + 3)}^{2} - 29$ $=(10x+3)^2-29$

$= {(10 x + 3)}^{2} - {(\sqrt{29})}^{2}$ $=(10x+3)^2-(sqrt(29))^2$

$= ((10 x + 3) - \sqrt{29}) ((10 x + 3) + \sqrt{29})$ $=((10x+3)-sqrt(29))((10x+3)+sqrt(29))$

$= (10 x + 3 - \sqrt{29}) (10 x + 3 + \sqrt{29})$ $=(10x+3-sqrt(29))(10x+3+sqrt(29))$

Hence $x = \frac{- 3 \pm \sqrt{29}}{10}$ $x = (-3+-sqrt(29))/10$

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How do you solve using the completing the square method $5 x^{2} + 3 x - 1 = 0$ $5x^2 + 3x - 1 = 0$ ?

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