How do you solve $2 x^{2} - 9 x + 4 = {(2 x - 1)}^{2}$ $2x^2-9x+4=(2x-1)^2$ ?

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How do you solve $2 x^{2} - 9 x + 4 = {(2 x - 1)}^{2}$ $2x^2-9x+4=(2x-1)^2$ ?

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Answer 1 · 2015-09-11T05:59:46

$x = - 3$ $x = -3$ or $x = \frac{1}{2}$ $x = 1/2$

Explanation:

$2 x^{2} - 9 x + 4 = {(2 x - 1)}^{2}$ $2x^2 - 9x + 4 = (2x - 1)^2$

We will start by expanding (FOILing ) the right-hand side of the equation:

$2 x^{2} - 9 x + 4 = 4 x^{2} - 4 x + 1$ $2x^2 - 9x + 4 = 4x^2 - 4x + 1$

Next, subtract $2 x^{2}$ $2x^2$ from both sides:

$- 9 x + 4 = 2 x^{2} - 4 x + 1$ $-9x + 4 = 2x^2 - 4x + 1$

Add $9 x$ $9x$ to both sides:

$4 = 2 x^{2} + 5 x + 1$ $4 = 2x^2 + 5x + 1$

And now, subtract 4 from both sides:

$0 = 2 x^{2} + 5 x - 3$ $0 = 2x^2 + 5x - 3$

This is a simple quadratic equation in $x$ $x$ so we can solve it by completing the square or applying the quadratic formula or whatever method you prefer. I'll apply the quadratic formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- 5 \pm \sqrt{5^{2} - 4 \cdot 2 \cdot (- 3)}}{2 \cdot 2}$ $x = (-b ± sqrt(b^2 - 4ac))/(2a) = (-5 ± sqrt(5^2 - 4*2*(-3)))/(2*2)$

Simplification gives us:

$x = \frac{- 5 \pm \sqrt{49}}{4}$ $x = (-5 ± sqrt(49))/4$

Which further reduces to:

$x = \frac{- 5 \pm 7}{4}$ $x = (-5 ± 7)/4$

From here it should be easy to see that

$x = - 3$ $x = -3$ or $x = \frac{1}{2}$ $x = 1/2$

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How do you solve $2 x^{2} - 9 x + 4 = {(2 x - 1)}^{2}$ $2x^2-9x+4=(2x-1)^2$ ?

1 Answer

Explanation:

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How do you solve $2 x^{2} - 9 x + 4 = {(2 x - 1)}^{2}$ $2x^2-9x+4=(2x-1)^2$ ?