Identification - HiSET

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Question

The equation

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

has two distinct solutions. What is their sum?

Answer

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 $ax^{2}+ bx + c = 0$ by subtracting from both sides:

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

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

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

If a quadratic equation has two distinct solutions, which we are given here, their sum is the linear coefficient . In this problem, , making the correct choice.

Compare your answer with the correct one above

Question

The graph of the polynomial function

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

has one and only one zero on the interval . On which subinterval is it located?

Answer

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

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

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

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

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

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

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

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

assumes positive values for and negative values for . By the IVT, has a zero on .

Compare your answer with the correct one above

Question

The equation

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

has two distinct solutions. What is their sum?

Answer

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 $ax^{2}+ bx + c = 0$ by subtracting from both sides:

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

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

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

If a quadratic equation has two distinct solutions, which we are given here, their sum is the linear coefficient . In this problem, , making the correct choice.

Compare your answer with the correct one above

Question

The graph of the polynomial function

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

has one and only one zero on the interval . On which subinterval is it located?

Answer

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

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

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

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

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

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

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

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

assumes positive values for and negative values for . By the IVT, has a zero on .

Compare your answer with the correct one above

Question

The equation

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

has two distinct solutions. What is their sum?

Answer

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 $ax^{2}+ bx + c = 0$ by subtracting from both sides:

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

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

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

If a quadratic equation has two distinct solutions, which we are given here, their sum is the linear coefficient . In this problem, , making the correct choice.

Compare your answer with the correct one above

Question

The graph of the polynomial function

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

has one and only one zero on the interval . On which subinterval is it located?

Answer

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

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

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

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

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

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

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

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

assumes positive values for and negative values for . By the IVT, has a zero on .

Compare your answer with the correct one above

Question

The equation

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

has two distinct solutions. What is their sum?

Answer

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 $ax^{2}+ bx + c = 0$ by subtracting from both sides:

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

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

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

If a quadratic equation has two distinct solutions, which we are given here, their sum is the linear coefficient . In this problem, , making the correct choice.

Compare your answer with the correct one above

Question

The graph of the polynomial function

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

has one and only one zero on the interval . On which subinterval is it located?

Answer

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

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

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

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

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

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

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

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

assumes positive values for and negative values for . By the IVT, has a zero on .

Compare your answer with the correct one above

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