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

Consider the equation:

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

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

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.

Consider the equation:

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

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

$- \frac{49}{12}$

$- \frac{49}{36}$

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.

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

$\frac{7}{3}$

$\frac{2}{3}$

$\frac{14}{15}$

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:

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

Solve for x: (x2 – x) / (x – 1) = 1

No solution

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 .

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 :

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

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

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.

Solve 3x2 + 10x = –3

x = –1/3 or –3

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.

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

-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