How do you solve $8x+1=4x^2$ by quadratic formula?

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How do you solve $8x+1=4x^2$ by quadratic formula?

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Answer 1 · 2016-02-07T08:17:11

$8x+1=4x^2$

$rarr8x=4x^2-1$

$rarr4x^2-1-8x=0$

In standard form:

$rarr4x^2-8x-1=0$

Now this is a Quadratic equation (in form $ax^2+bx^2+c=0$ )

Use Quadratic formula:

$x=(-b+-sqrt(b^2-4ac))/(2a)$

In this case $a=4,b=-8,c=-1$

Substitute the values into the formula:

$rarrx=(-(-8)+-sqrt((-8)^2-4(4)(-1)))/(2(4))$

$rarrx=(8+-sqrt(64-(-16)))/8$

$rarrx=(8+-sqrt(64+16))/8$

$rarrx=(8+-sqrt(80))/8$

$rarrx=(8+-sqrt(16*5))/8$

$rarrx=(8+-4sqrt5)/8$

$rarrx=(2+-sqrt5)/2$

Answer 2 · 2016-02-07T08:24:14

$(2+-sqrt(5))/(2)$

Explanation:

quadratic fomula

$ax^2+bx+c=0$

$(-b+-sqrt(b^2 -4ac))/(2a)$

so $8x+1=4x^2$

$rarr4x^2-8x-1=0$

$rarr(-(-8)+-sqrt((-8)^2 -4(4)(-1)))/(2(4))$

$rarr(8+-sqrt(64 +16))/(8)$

$rarr(8+-sqrt(80))/(8)$

$rarr(8+-4sqrt(5))/(8)$

$rarr(2+-sqrt(5))/(2)$

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