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HiSET: Math : Identification

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Example Question #1 : Identification

The equation

$x^{2} + 34 = 17x$ $\displaystyle x^{2} + 34 = 17x$

has two distinct solutions. What is their sum?

Possible Answers:

$\displaystyle -17$

$\displaystyle -34$

$\displaystyle 17$

$\displaystyle 34$

$\displaystyle 0$

Correct answer:

$\displaystyle -17$

Explanation:

It is not necessary to actually find the solutions to a quadratic equation to determine the sum of its solutions.

First, get the equation in standard form $ax^{2}+ bx + c = 0$ $\displaystyle ax^{2}+ bx + c = 0$ by subtracting 17x $\displaystyle 17x$ from both sides:

$x^{2} + 34 = 17x$ $\displaystyle x^{2} + 34 = 17x$

$x^{2} + 34 -17x = 17x - 17x$ $\displaystyle x^{2} + 34 -17x = 17x - 17x$

$x^{2} -17x + 34 = 0$ $\displaystyle x^{2} -17x + 34 = 0$

If a quadratic equation has two distinct solutions, which we are given here, their sum is the linear coefficient $\displaystyle b$. In this problem, b= -17 $\displaystyle b= -17$, making $\displaystyle -17$ the correct choice.

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Example Question #44 : Algebraic Concepts

The graph of the polynomial function

$f(x)= x^{3}+x^{2}-5x+2$ $\displaystyle f(x)= x^{3}+x^{2}-5x+2$

has one and only one zero on the interval (0, 1) $\displaystyle (0, 1)$. On which subinterval is it located?

Possible Answers:

$\displaystyle (0.8, 1)$

$\displaystyle (0.2, 0.4)$

$\displaystyle (0.4, 0.6)$

$\displaystyle (0, 0.2)$

$\displaystyle (0,6, 0.8)$

Correct answer:

$\displaystyle (0.4, 0.6)$

Explanation:

The Intermediate Value Theorem (IVT) states that if the graph of a function $\displaystyle f$ is continuous on an interval [a, b] $\displaystyle [a, b]$, and f(a) $\displaystyle f(a)$ and f(b) $\displaystyle f(b)$ differ in sign, then $\displaystyle f$ has a zero on (a,b) $\displaystyle (a,b)$. Consequently, the way to answer this question is to determine the signs of $\displaystyle f$ on the endpoints of the subintervals - $\left \{ 0, 0.2, 0.4, 0.6, 0.8, 1 \right \}$ $\displaystyle \left \{ 0, 0.2, 0.4, 0.6, 0.8, 1 \right \}$. We can do this by substituting each value for $\displaystyle x$ as follows:

$f(x)= x^{3}+x^{2}-5x+2$ $\displaystyle f(x)= x^{3}+x^{2}-5x+2$

$f(0)= 0^{3}+0^{2}-5(0)+2 = 0+0-0+2= 2$ $\displaystyle f(0)= 0^{3}+0^{2}-5(0)+2 = 0+0-0+2= 2$

$f(0.2)= 0.2^{3}+0.2^{2}-5(0.2)+2 = 0.008+0.04-1+2 = 1.048$ $\displaystyle f(0.2)= 0.2^{3}+0.2^{2}-5(0.2)+2 = 0.008+0.04-1+2 = 1.048$

$f(0.4)= 0.4^{3}+0.4^{2}-5(0.4)+2= 0.064+ 0.16- 2+2 = 0.224$ $\displaystyle f(0.4)= 0.4^{3}+0.4^{2}-5(0.4)+2= 0.064+ 0.16- 2+2 = 0.224$

$f(0.6)= 0.6^{3}+0.6^{2}-5(0.6)+2= 0.216+ 0.36- 3+2 = -0.424$ $\displaystyle f(0.6)= 0.6^{3}+0.6^{2}-5(0.6)+2= 0.216+ 0.36- 3+2 = -0.424$

$f(0.8)= 0.8^{3}+0.8^{2}-5(0.8)+2= 0.512+ 0.64- 4+2 = -0.848$ $\displaystyle f(0.8)= 0.8^{3}+0.8^{2}-5(0.8)+2= 0.512+ 0.64- 4+2 = -0.848$

$f(1)=1^{3}+1^{2}-5(1)+2 =1+1-5+2=-1$ $\displaystyle f(1)=1^{3}+1^{2}-5(1)+2 =1+1-5+2=-1$

$\displaystyle f$ assumes positive values for $\displaystyle x= 0, 0.2, 0.4$ and negative values for $\displaystyle x = 0.6, 0.8, 1$. By the IVT, $\displaystyle f$ has a zero on $\displaystyle (0.4, 0.6)$.

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HiSET: Math : Identification

Study concepts, example questions & explanations for HiSET: Math

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Example Question #1 : Identification

Example Question #44 : Algebraic Concepts

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