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How do you use the quadratic formula to find both solutions to the quadratic equation $x^{2} + 3 x - 4 = 0$ $x^2+3x-4=0$ ?

2 Answers

Jun 29, 2015

By replacing the variable coefficients in the general quadratic formual with the coefficients for the terms in the given equation we can get $x = 1$ $x=1$ or $x = - 4$ $x=-4$

Explanation:

For an equation in the form:
$XXXX$ $color(white)("XXXX")$ $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$
the quadratic formula gives the solutions as
$XXXX$ $color(white)("XXXX")$ $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x= (-b+-sqrt(b^2-4ac))/(2a)$

$x^{2} + 3 x - 4 = 0$ $x^2+3x-4 = 0$ is in this form with
$XXXX$ $color(white)("XXXX")$ $a = 1$ $a=1$
$XXXX$ $color(white)("XXXX")$ $b = 3$ $b=3$ and
$XXXX$ $color(white)("XXXX")$ $c = (- 4)$ $c=(-4)$

Replacing $a, b, and c$ $a, b, and c$ in the quadratic formula, we get
$XXXX$ $color(white)("XXXX")$ $x = \frac{- 3 \pm \sqrt{3^{2} - 4 (1) (- 4)}}{2 (1)}$ $x=(-3+-sqrt(3^2-4(1)(-4)))/(2(1))$

$XXXX$ $color(white)("XXXX")$ $x = \frac{- 3 \pm \sqrt{9 + 16}}{2}$ $x= (-3+-sqrt(9+16))/2$

$XXXX$ $color(white)("XXXX")$ $x = \frac{- 3 \pm 5}{2}$ $x=(-3+-5)/2$

$XXXX$ $color(white)("XXXX")$ $x = \frac{2}{2}$ $x = 2/2$ or $x = \frac{- 8}{2}$ $x= (-8)/2$

$x = 1$ $x=1$ or $x = - 4$ $x=-4$

Luca F. · George C.

Jun 29, 2015

$x_{1} = - 4$ $x_1=-4$
$x_{2} = 1$ $x_2=1$

Explanation:

Having the equation $x^{2} + 3 x - 4 = 0$ $x^2+3x−4=0$ , we find the $Delta_x$ :
$Delta_x=3^2-4*1*(-4)=9+16=25$

Now we find $x_1$ and $x_2$ with the quadratic formula .

$x_"1,2"=(-3+-sqrt(Delta_x))/(2*1)=(-3+-5)/2$ .

The two solutions are:

$x_1=(-3-5)/2=-4$

$x_2=(-3+5)/2=1$ .

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