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PSAT Math : How to find the solution to a quadratic equation

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Example Questions

Example Question #23 : How To Find The Solution To A Quadratic Equation

The formula to solve a quadratic expression is:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

All of the following equations have real solutions EXCEPT:

Possible Answers:

2x^2-4x+5=0

-2x^2+4x+5=0

2x^2-4x=0

2x^2-4x-5=0

-2x^2-4x+5=0

Correct answer:

2x^2-4x+5=0

Explanation:

We can use the quadratic formula to find the solutions to quadratic equations in the form ax²+ bx + c = 0. The quadratic formula is given below.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

In order for the formula to give us real solutions, the value under the square root, b²– 4ac, must be greater than or equal to zero. Otherwise, the formula will require us to find the square root of a negative number, which gives an imaginary (non-real) result.

In other words, we need to look at each equation and determine the value of b²–4ac. If the value of b²– 4ac is negative, then this equation will not have real solutions.

Let's look at the equation 2x² – 4x + 5 = 0 and determine the value of b²– 4ac.

b²– 4ac = (–4)² – 4(2)(5) = 16 – 40 = –24 < 0

Because the value of b²– 4ac is less than zero, this equation will not have real solutions. Our answer will be 2x² – 4x + 5 = 0.

If we inspect all of the other answer choices, we will find positive values for b²– 4ac, and thus these other equations will have real solutions.

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Example Question #24 : How To Find The Solution To A Quadratic Equation

Let f(x)=x^2-4x+2 , and let g(x)=6-x . What is the sum of the possible values of such that f(k)=g(k) .

Possible Answers:

Correct answer:

Explanation:

We are told that f(x) = x² - 4x + 2, and g(x) = 6 - x. Let's find expressions for f(k) and g(k).

f(k) = k² – 4k + 2

g(k) = 6 – k

Now, we can set these expressions equal.

f(k) = g(k)

k² – 4k +2 = 6 – k

Add k to both sides.

k² – 3k + 2 = 6

Then subtract 6 from both sides.

k² – 3k – 4 = 0

Factor the quadratic equation. We must think of two numbers that multiply to give us -4 and that add to give us -3. These two numbers are –4 and 1.

(k – 4)(k + 1) = 0

Now we set each factor equal to 0 and solve for k.

k – 4 = 0

k = 4

k + 1 = 0

k = –1

The two possible values of k are -1 and 4. The question asks us to find their sum, which is 3.

The answer is 3.

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Example Question #21 : How To Find The Solution To A Quadratic Equation

Stuff

Note: Figure NOT drawn to scale.

Refer to the above diagram, which shows Rectangle RECT with RT = 10, RE = 20 .

is the midpoint of $\overline{EC}$ ; $\Delta ART \sim \Delta NEA$ ; AR < AE

Evaluate (to the nearest tenth, if applicable).

Possible Answers:

5.0

2.9

3.5

Insufficient information is given to answer the question.

2.2

Correct answer:

2.9

Explanation:

The corresponding sides of similar triangles are in proportion, so we can set up and solve the proportion statement for :

$\Delta ART \sim \Delta NEA$ , so

$\frac{AR}{NE} = \frac{RT}{EA}$

For the sake of simplicuty, we will let x = AR

Since is the midpoint of $\overline{EC}$ , $NE = \frac{1}{2} EC =\frac{1}{2} RT = \frac{1}{2} \cdot 10 = 5$ .

Also, EA = RA - AR = 20 - AR = 20 -x .

The proportion statement becomes

$\frac{x}{5} = \frac{10}{20-x}$

Solve for using cross-products:

$x \cdot (20-x ) = 5 \cdot 10$

$20x-x ^{2} = 5 0$

$x ^{2} -20x + 5 0 = 0$

By the quadratic equation, setting a = 1, b = -20, c = 50 :

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$x = \frac{-(-20) \pm \sqrt{(-20)^{2}-4 \cdot 1 \cdot 50}}{2 \cdot 1}$

$x = \frac{20 \pm \sqrt{400-200}}{2 }$

$x = \frac{20 \pm \sqrt{ 200}}{2 }$

$x = \frac{20 \pm 14.14}{2 }$

There are two possibilities:

$x = \frac{20 + 14.14}{2 } \approx 17.1$

$x = \frac{20 - 14.14}{2 } \approx 2.9$

$\overline{RE}$ is divided into segments of length 2.9 and 17.1. The lesser is the length of $\overline{AR}$ , so the correct choice is 2.9.

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Example Question #21 : How To Find The Solution To A Quadratic Equation

Solve for :

$4x^{2} + 9=3(x^{2}+4)$

Possible Answers:

$\pm \sqrt{3}$

$\frac{3}{2}$

$\frac{3}{4}$

Correct answer:

$\pm \sqrt{3}$

Explanation:

Begin by distributing the three on the right side of the equation: $4x^{2}+9=3x^{2}+12$

Next combine your like terms by subtracting $3x^{2}$ from both sides to give you $x^{2}+9=12$

Next, subtract 9 from both sides to give you $x^{2}=3$ . To solve for , now take the square root of both sides. This gives you the answer, $x=\pm \sqrt{3}$

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PSAT Math : How to find the solution to a quadratic equation

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #23 : How To Find The Solution To A Quadratic Equation

Example Question #24 : How To Find The Solution To A Quadratic Equation

Example Question #21 : How To Find The Solution To A Quadratic Equation

Example Question #21 : How To Find The Solution To A Quadratic Equation

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