Question

Use the quadratic formula to find both solutions to the quadratic equation given below 4x^2+3x-1=0

Answer 1

    4x²  +  3x    -  1 = 0
a=4      b=3   c=-1

Quadratic formula:
x =$\frac{-b +/- \sqrt{b^{2} -4ac} }{2a} = \frac{-3 +/- \sqrt{3^{2} - 4(4)(-1)}}{2(4)}$
= $\frac{-3 +/- \sqrt{9 + 16}}{8} = \frac{-3 +/- \sqrt{25}}{8} = \frac{-3 +/- 5}{8}$
= $\frac{-3 + 5}{8}$ , $\frac{-3 - 5}{8}$
= $\frac{2}{8}$, $\frac{-8}{8}$
= $\frac{1}{4}$, -1

Answer 2

The solutions to the quadratic equation are:

x = 1/4 and x = -1.

Quadratic equation formula

Given:

$4 x^{2}+3 x-1=0$

Solve with the quadratic equation formula

$x=\frac{-b \pm \sqrt{b^{2}-4 \cdot ac}}{2 \cdot a}$

Let a = 4, b = 3 and c = -1

Substituting the values of a, b, and c in the above equation, we get

$x_{1,2}=\frac{-3 \pm \sqrt{3^{2}-4 \cdot 4(-1)}}{2 \cdot 4}$

$= \sqrt{3^{2}-4 \cdot 4(-1)}=5$

$x_{1,2}=$ $-3 \pm \frac{5}{2*4}$

Separate the solutions

$x_{1}=\frac{-3+5}{2 \cdot 4}$ and $x_{2}=\frac{-3-5}{2 \cdot 4}$

then we get

$x=\frac{-3+5}{2 \cdot 4}$

x = 1/4 and

$x=\frac{-3-5}{2 \cdot 4}$

x = -1

The solutions to the quadratic equation are:

x = 1/4 and x = -1.

Therefore, the values of x are 1/4 and -1.

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