How do you solve $x^2 + 3x - 4 = 0$ graphically and algebraically?

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Answer 1 · 2016-07-22T10:01:35

$x=1$
$x=-4$

Given -

$x^2+3x-4=0$

Algebraic Solution

$x^2-x+4x-4=0$
$x(x-1)+4(x-1)=0$
$(x-1)(x+4)=0$
$x=1$
$x=-4$

If you take the above equation as a function as

$y=x^2+3x-4$

One Solution is

$(1, 0)$

Another solution is

$(-4,0)$

To have a graphic solution, You have identify a range of

values for $x$ That includes vertex and the above said two points.

Its vertex is given by -

$x=(-b)/(2xxa)=(-3)/(2 xx1)=-3/2$

Now take four values above and below $-3/2$

Find the corresponding $y$ values for $x$

Plot all the values. Those co-ordinates where the curve cuts the x-axis is the graphic solution to the problem.

Look at the image for solution

How do you solve x^2 + 3x - 4 = 0x^2 + 3x - 4 = 0 graphically and algebraically?