Abe come is paying $400 a month to rent a house

Mathematics

1 answer:

tatyana61 [14]3 years ago

7 0

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What are the zeros of the quadratic function f(x) = 2x2 + 8x – 3?

jenyasd209 [6]

Answer:

$x=-2-\sqrt{\dfrac{11}{2}}\ \text{and}\ x=-2+\sqrt{\dfrac{11}{2}}$

Step-by-step explanation:

My favorite way to go at this is to look at a graph. It shows the vertex at (-2, -11). Since the leading coefficient is 2, this means the roots are ...

$-2\pm\sqrt{\dfrac{11}{2}}$

where the 2 in the denominator of the radical is the leading coefficient.

You can also use other clues:

the axis of symmetry is -b/(2a) = -8/(2(2)) = -2, so answer choices C and D don't work
the single change in sign in the coefficients (+ + -) tells you there is one positive real root, so answer choice B doesn't work.

The first answer choice is the only one with values symmetrical about -2 and one of them positive.

You may be expected to use the quadratic formula:

$x=\dfrac{-b\pm\sqrt{b^-4ac}}{2a}=\dfrac{-8\pm\sqrt{8^2-4(2)(-3)}}{2(2)}\\\\=\dfrac{-8}{4}\pm\dfrac{\sqrt{88}}{4}=-2\pm\sqrt{\dfrac{11}{2}}$