Math

QuestionDetermine the zeros of the function f(x)=2x2+10x+11f(x)=2x^{2}+10x+11 and identify the xx-intercepts. Choose A, B, or C.

Studdy Solution

STEP 1

Assumptions1. The function is a quadratic function given by f(x)=x+10x+11f(x)= x^{}+10 x+11 . We are looking for the zeros of the function, which are also the xx-intercepts of the graph of the function.

STEP 2

To find the zeros of the function, we need to set the function equal to zero and solve for xx.
0=2x2+10x+110 =2x^{2} +10x +11

STEP 3

This is a quadratic equation in the form ax2+bx+c=0ax^{2} + bx + c =0. We can solve it using the quadratic formula, which is given byx=b±b2ac2ax = \frac{-b \pm \sqrt{b^{2} -ac}}{2a}

STEP 4

Now, plug in the values for aa, bb, and cc from our quadratic equation into the quadratic formula.
x=10±(10)2421122x = \frac{-10 \pm \sqrt{(10)^{2} -4*2*11}}{2*2}

STEP 5

implify the expression under the square root.
x=10±100884x = \frac{-10 \pm \sqrt{100 -88}}{4}

STEP 6

Calculate the value under the square root.
x=10±124x = \frac{-10 \pm \sqrt{12}}{4}

STEP 7

implify the square root.
x=10±234x = \frac{-10 \pm2\sqrt{3}}{4}

STEP 8

Finally, divide each term in the numerator by the denominator to get the zeros of the function.
x=104±234x = -\frac{10}{4} \pm \frac{2\sqrt{3}}{4}

STEP 9

implify the fractions.
x=52±32x = -\frac{5}{2} \pm \frac{\sqrt{3}}{2}Therefore, the zeros of the function are x=52+32x = -\frac{5}{2} + \frac{\sqrt{3}}{2} and x=5232x = -\frac{5}{2} - \frac{\sqrt{3}}{2}. These are also the xx-intercepts of the graph of the function.So, the correct choice is B. The zeros and the xx-intercepts are the same. They are x=52+32x = -\frac{5}{2} + \frac{\sqrt{3}}{2} and x=5232x = -\frac{5}{2} - \frac{\sqrt{3}}{2}.

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