How do you solve $2 x^{2} + 5 x = 3$ $2x^2 +5x=3$ ?

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Answer 1 · 2016-04-05T09:52:21

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

Given: $2 x^{2} + 5 x = 3$ $" "2x^2+5x=3$
Write as: $2 x^{2} + 5 x - 3 = 0$ $" "2x^2+5x-3=0$

The whole number factors of 3 are 1 and 3.
The whole number factors of 2 are 1 and 2.

Attempt 1

$(2 x + 3) (x + 1) = 2 x^{2} + 2 x + 3 x + 3 Fail$ $(2x+3)(x+1) = 2x^2+2x+3x+3color(red)(" Fail")$

Attempt 2

$(2 x + 1) (x - 3) = 2 x^{2} - 6 x + x - 3 Fail$ $(2x+1)(x-3) = 2x^2-6x+x-3color(red)(" Fail")$

Looks as though there is at least 1 non whole number solution to this so use the formula

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

$x = \frac{- 5 \pm \sqrt{5^{2} - 4 (2) (- 3)}}{2 (2)}$ $x=(-5+-sqrt(5^2-4(2)(-3)))/(2(2))$

$x = \frac{- 5 \pm \sqrt{25 + 24}}{4}$ $x=(-5+-sqrt(25+24))/4$

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

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

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

How do you solve 2x^2 +5x=32x2+5x=32x^2 +5x=3?