Question

Will the graph of the equation above intersect the x-axis in zero, one, or two points?

Answer 1

Answer: Option D

Step-by-step explanation:

Add 4x² to both sides  of the equation:

$16x+8+4x^2=-4x^2+4x^2\\4x^2+16x+8=0$

The Quadratic formula is:

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

You can identify that in this case:

$a=4\\b=16\\c=8$

Substitute into the Quadratic formula:

$x=\frac{-16\±\sqrt{(16)^2-4(4)(8)}}{2(4)}$

 $x_1=-0.58\\x_2=-3.41$

Therefore, the graph of the equation above intersect the x-axis in two points.

Answer 2

Hello!

The answer is:

D. The parabola intercepts the x-axis in two points.

$x_{1}=-0.59\\\\x_{2}=-3.41$

Why?

To find the parabola intercepts, we need to use the quadratic formula. The parabola intercepts with the x-axis, are also called roots or zeroes.

The quadratic formula states that:

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

We are given the parabola:

$16x+8=-4x^{2}$

Which is also equal to:

$4x^{2}+16x+8=0$

Where,

$a=4\\b=16\\c=8$

Then, substituting into the quadratic formula to find the roots of the parabola, we have:

$x=\frac{-b+-\sqrt{b^{2}-4ac } }{2a}\\\\x=\frac{-16+-\sqrt{(16)^{2}-4*4*8 } }{2*4}=\frac{-16+-\sqrt{(256-128) } }{8}\\\\x==\frac{-16+-\sqrt{(256-128) } }{8}=\frac{-16+-\sqrt{(28 } }{8}\\\\x=\frac{-16+-\sqrt{128 } }{8}=\frac{-16+-(11.31) }{8}\\\\\\x_{1}=\frac{-16+(11.31) }{8}=-0.59\\\\x_{1}=\frac{-16-(11.31) }{8}=-3.41$

Hence,the parabola intercepts the x-axis in two points.

$x_{1}=-0.59\\\\x_{2}=-3.41$

Have a nice day!