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Softa [21]
2 years ago
11

Determine which functions have two real number zeros by calculating the discriminant, b2 – 4ac. Check all that apply. Please ans

wer f(x) = x2 + 6x + 8 g(x) = x2 + 4x + 8 h(x) = x2 – 12x + 32 k(x) = x2 + 4x – 1 p(x) = 5x2 + 5x + 4 t(x) = x2 – 2x – 15
Mathematics
2 answers:
timama [110]2 years ago
4 0
<span>The discriminant of a quadratic equation is the b^2-4ac portion that the square root is taken of. If the discriminant is negative, then the function has 2 imaginary roots, if the discriminant is equal to 0, then the function has only 1 real root, and finally, if the discriminant is greater than 0, the function has 2 real roots. So let's look at the equations and see which have a positive discriminant. f(x) = x^2 + 6x + 8 6^2 - 4*1*8 36 - 32 = 4 Positive, so f(x) has 2 real roots. g(x) = x^2 + 4x + 8 4^2 - 4*1*8 16 - 32 = -16 Negative, so g(x) does not have any real roots h(x) = x^2 – 12x + 32 -12^2 - 4*1*32 144 - 128 = 16 Positive, so h(x) has 2 real roots. k(x) = x^2 + 4x – 1 4^2 - 4*1*(-1) 16 - (-4) = 20 Positive, so k(x) has 2 real roots. p(x) = 5x^2 + 5x + 4 5^2 - 4*5*4 25 - 80 = -55 Negative, so p(x) does not have any real roots t(x) = x^2 – 2x – 15 -2^2 - 4*1*(-15) 4 - (-60) = 64 Positive, so t(x) has 2 real roots.</span>
Korvikt [17]2 years ago
4 0

Answer:

Answers are A, C, D, & F

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Step-by-step explanation:

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cos(\theta)=\frac{\sqrt{2}}{2}

\frac{3\pi}{2}

so

The angle \theta belong to the third or fourth quadrant

The value of sin(\theta) is negative

Step 1

Find the value of  sin(\theta)

Remember

sin^{2}(\theta)+cos^{2}(\theta)=1

we have

cos(\theta)=\frac{\sqrt{2}}{2}

substitute

sin^{2}(\theta)+(\frac{\sqrt{2}}{2})^{2}=1

sin^{2}(\theta)=1-\frac{1}{2}

sin^{2}(\theta)=\frac{1}{2}

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Step 2

Find the value of tan(\theta)

tan(\theta)=\frac{sin(\theta)}{cos(\theta)}

we have

sin(\theta)=-\frac{\sqrt{2}}{2}

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substitute

tan(\theta)=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}

tan(\theta)=-1

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