Solve for xx 8x2−20x+12=48x^2-20x+12=4

x = 2 , x = 1 2 x=2,x=\frac12 x = 2 , x = 2 1

Solve for x x 8 x 2 − 20 x + 12 = 4 8x^2-20x+12=4

Here, we want to solve this equation for x. So to do that, we need to get x all by itself. So we're going to complete the square and then use our square root property to do just that. So first, let's move our constant four over to the left hand side so we can get our quadratic in standard form to complete the square. We have 12 minuteus four, which is eight, and all of that is now equal to zero. Now since we have an a value of something other than one, we're going to divide each of the terms in this equation by that value. So we're dividing everything by eight and we now have x squared minus 20 over eight reduces to five halves x plus one is equal to zero. And now we complete the square. So drawing out our square diagram, we're going to represent our x squared term here as the side lengths x times x. And we'll take our negative five halves x and split that in half to get negative five fourths x and negative five fourths x. And since these side lengths are already x, that means our other side lengths here are negative five fourths. And to complete this square, we need to multiply negative five fourths by itself, which gets us 25 sixteenths. So we can closely represent this quadratic with the perfectly square trinomial of x minus five fourths square, which is equal to x squared minus five halves x plus 25 sixteenths. But of course we don't want 25 sixteenths, we want positive one. So what do I need to add or subtract to 25 sixteenths to get one? Well, that's going to be the difference between twenty five and sixteen here, which is nine sixteenths. So we want to subtract nine sixteenths from both sides. And now we have x minus five fourths squared minus nine sixteenths, and that's going to equal x squared minus five halves x plus 25 sixteenths minus nine sixteenths is sixteenths sixteenths, which is one. So now that we know how to represent our quadratic here in this form, we can use that form and set it equal to zero and solve for x. So we have x minus five fourths squared minus nine sixteenths equal to zero. And we'll add nine sixteenths to both sides to get x minus fivefour squared is equal to nine sixteenths. And now we can use our square root property and square root both sides. We have x minus fivefour is equal to positive and negative. Square root of nine is three and square root of 16 is four. Next we'll add five fourths to both sides. And we have x is equal to five fourths plus and minus three fourths. So simplifying this, we have five fourths plus three fourths, which gives us eight fourths, which simplifies to x equaling two as one of our solutions. And we'd have x equals five fourths minus three fourths, which is two fourths, which simplifies to x equals one half as our second solution.

Solve for x x 8 x 2 − 20 x + 12 = 4 8x^2-20x+12=4

x = 2 , x = 1 2 x=2,x=\frac12 x = 2 , x = 2 1

x = 3 , x = − 1 2 x=3,x=-\frac12 x = 3 , x = − 2 1

x = 3 , x = 1 2 x=3,x=\frac12 x = 3 , x = 2 1

x = 2 , x = − 1 2 x=2,x=-\frac12 x = 2 , x = − 2 1

10. Quadratic Equations and Functions

Completing the Square

Intermediate Algebra

10. Quadratic Equations and Functions

Completing the Square

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Multiple Choice

Solve for $x$
$8 x^{2} - 20 x + 12 = 4$

$x=3,x=-\frac12$

$x=3,x=\frac12$

$x=2,x=-\frac12$

$x=2,x=\frac12$

Verified step by step guidance

Start with the given quadratic equation: \$8x^2 - 20x + 12 = 4$.

Move all terms to one side to set the equation equal to zero: \$8x^2 - 20x + 12 - 4 = 0$, which simplifies to $8x^2 - 20x + 8 = 0$.

Divide the entire equation by the greatest common factor (GCF) of the coefficients to simplify. The GCF of 8, -20, and 8 is 4, so divide each term by 4 to get \$2x^2 - 5x + 2 = 0$.

Use the quadratic formula to solve for $x$. Recall the quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=2$, $b=-5$, and $c=2$.

Calculate the discriminant $b^2 - 4ac$, then find the two possible values for $x$ by substituting into the quadratic formula. These two values will be the solutions to the equation.

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