How to find the solution to a quadratic equation

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SAT Math › How to find the solution to a quadratic equation

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Consider the equation:

$3 x^{2} + 48 =$ _________

Fill in the blank with a real constant to form an equation with exactly one real solution.

CORRECT

Explanation

We will call the constant that goes in the blank . The equation becomes

$3 x^{2} + 48 = Nx$

Write the quadratic equation in standard form $ax^{2} + bx+ c = 0$ by subtracting from both sides:

$3 x^{2} + 48 - Nx = Nx - Nx$

$3 x^{2} - Nx + 48 = 0$

The solution set comprises exactly one rational solution if and only if the discriminant $b^{2} - 4ac$ is equal to 0. Setting . and substituting in the equation:

$b^{2} - 4ac = 0$

$(-N)^{2} - 4 (3)(48)= 0$

$N^{2} - 576= 0$

Solving for :

$N^{2} - 576+ 576 = 0+ 576$

$N^{2} = 576$

$N = \pm \sqrt{576}$

$N = \pm 24$ ,

that is, either or .

is not a choice, but 24 is; this is the correct response.

What is the sum of all the values of that satisfy:

$\frac{7}{3}$

CORRECT

$\frac{2}{3}$

$\frac{14}{15}$

Explanation

With quadratic equations, always begin by getting it into standard form:

Therefore, take our equation:

And rewrite it as:

You could use the quadratic formula to solve this problem. However, it is possible to factor this if you are careful. Factored, the equation can be rewritten as:

Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :

$x=\frac{-2}{3}$

The sum of these values is:

$3+(-\frac{2}{3})=3-\frac{2}{3}=\frac{9}{3}-\frac{2}{3}=\frac{7}{3}$

Consider the equation:

$3x^{2} + 7x =$ __________

Fill in the blank with a real constant to form an equation with exactly one real solution.

$- \frac{49}{12}$

CORRECT

$- \frac{49}{36}$

None of the other responses gives a correct answer.

Explanation

We will call the constant that goes in the blank . The equation becomes

$3x^{2} + 7x = N$

Write the quadratic equation in standard form $ax^{2} + bx+ c = 0$ by subtracting from both sides:

$3x^{2} + 7x - N = N - N$

$3x^{2} + 7x - N = 0$

The solution set comprises exactly one rational solution if and only if the discriminant $b^{2} - 4ac$ is equal to 0. Setting . and substituting in the equation:

$b^{2} - 4ac = 0$

$7^{2} - 4(3)(-N) = 0$

Solving for :

$12N\div 12 = -49 \div 12$

$N = - \frac{49}{12}$ ,

the correct response.

Solve for x: (x2 – x) / (x – 1) = 1

No solution

CORRECT

x = 1

x = -1

x = 2

x = -2

Explanation

Begin by multiplying both sides by (x – 1):

x2 – x = x – 1

Solve as a quadratic equation: x2 – 2x + 1 = 0

Factor the left: (x – 1)(x – 1) = 0

Therefore, x = 1.

However, notice that in the original equation, a value of 1 for x would place a 0 in the denominator. Therefore, there is no solution.

Evaluate .

CORRECT

The system has no solution.

Explanation

Multiply both sides of the top equation by 7:

Multiply both sides of the bottom equation by :

Add both sides of the equations to eliminate the terms:

$\underline{-1.6x-2.8y = 3.8}$

Solve for :

$3.3x \div 3.3 = 4.29 \div 3.3$

Solve 3x2 + 10x = –3

x = –1/3 or –3

CORRECT

x = –1/6 or –6

x = –1/9 or –9

x = –2/3 or –2

x = –4/3 or –1

Explanation

Generally, quadratic equations have two answers.

First, the equations must be put in standard form: 3x2 + 10x + 3 = 0

Second, try to factor the quadratic; however, if that is not possible use the quadratic formula.

Third, check the answer by plugging the answers back into the original equation.

If $x(x^2-5x+3)=-9, \ x>0,\ and \ y^2+y=2$ then which of the following is a possible value for ?

CORRECT

Explanation

$x(x^2-5x+3)=-9, \ x>0,\ and \ y^2+y=2$

$x(x^2-5x+3)=-9 \Rightarrow x^3-5x^2+3x + 9=0\Rightarrow (x^2-6x+9)(x+1)$

$(x^2-6x+9)(x+1) = (x-3)^2(x+1)\Rightarrow x= -1\ or\ 3$

Since , .

$y^2+y-2=0 \Rightarrow (y+2)(y-1)=0 \Rightarrow y=-2\ or\ 1.$

Thus $xy^2+y = (3)(-2)^2+(-2)\ or\ (3)(1)^2+(1) = 12-2\ or\ 3+1\ = 10\ or\ 4$

Of these two, only 4 is a possible answer.

The expression $x^{2} - 8x +12$ is equal to 0 when $x = 2$ and $x = ?$

$6$

CORRECT

$-12$

$-6$

$-2$

$4$

Explanation

Factor the expression and set each factor equal to 0:

$(x-2)(x-6)= 0$

$x-2 = 0$

$x = 2$

$x-6 = 0$

$x = 6$

Consider the equation

$4 x ^{2}+ 21 = 5x$ .

Which of the following statements correctly describes its solution set?

Exactly two solutions, both of which are imaginary.

CORRECT

Exactly two solutions, both of which are irrational.

Exactly two solutions, both of which are rational.

Exactly one solution, which is rational.

Exactly one solution, which is irrational.

Explanation

Write the quadratic equation in standard form $ax^{2} + bx+ c = 0$ by subtracting from both sides:

$4 x ^{2}+ 21 = 5x$

$4 x ^{2}+ 21 - 5x = 5x - 5x$

$4 x ^{2}- 5x + 21 = 0$

The nature of the solution set of a quadratic equation in standard form can be determined by examining the discriminant $b^{2} - 4ac$ . Setting :

$b^{2} - 4ac =(-5) ^{2} - 4 \cdot 4 \cdot 21 = 25 - 336= -311$

The discriminant is negative, so there are two imaginary solutions.

Let f(x) = 2_x_2 – 4_x_ + 1 and g(x) = (_x_2 + 16)(1/2). If k is a negative number such that f(k) = 31, then what is the value of (f(g(k))?

-81

CORRECT

-35

Explanation

In order to find the value of f(g(k)), we will first need to find k. We are told that f(k) = 31, so we can write an expression for f(k) and solve for k.

f(x) = 2_x_2 – 4_x_ + 1

f(k) = 2_k_2 – 4_k_ + 1 = 31

Subtract 31 from both sides.

2_k_2 – 4_k –_ 30 = 0

Divide both sides by 2.

k_2 – 2_k – 15 = 0

Now, we can factor this by thinking of two numbers that multiply to give –15 and add to give –2. These two numbers are –5 and 3.

k_2 –2_k – 15 = (k – 5)(k + 3) = 0

We can set each factor equal to 0 to find the values for k.

k – 5 = 0

Add 5 to both sides.

k = 5