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

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Example Question #31 : How To Find The Solution To A Quadratic Equation

Consider the equation:

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

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

Possible Answers:

$- \frac{49}{36}$

None of the other responses gives a correct answer.

$- \frac{49}{12}$

Correct answer:

$- \frac{49}{12}$

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 a = 3, b= 7 , c= -N . and substituting in the equation:

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

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

Solving for :

49 - (-12N) = 0

49 + 12 N = 0

49 + 12 N- 49 = 0 - 49

12N = -49

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

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

the correct response.

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Example Question #32 : How To Find The Solution To A Quadratic Equation

Consider the equation:

__________ $x^{2} + 16 = 7x$

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

Possible Answers:

None of the other responses gives a correct answer.

$\frac{7}{8}$

$\frac{49}{64}$

$\frac{64}{49}$

$\frac{8}{7}$

Correct answer:

$\frac{49}{64}$

Explanation:

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

$N x^{2} + 16 = 7x$

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

$N x^{2} + 16 - 7x = 7x - 7x$

$N x^{2}- 7x + 16 = 0$

The solution set comprises exactly one rational solution if and only if the discriminant $b^{2} - 4ac$ is equal to 0. Setting a =N, b= -7 , c= 16 . and substituting in the equation:

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

$(-7)^{2} - 4 (N)(16)= 0$

49 - 64N= 0

Solving for :

49 - 64N+ 64N = 0 + 64N

49 = 64N

$49 \div 64 = 64N \div 64$

$N = \frac{49}{64}$ ,

the correct response.

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Example Question #2121 : Sat Mathematics

Consider the equation:

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

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

Possible Answers:

Correct answer:

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 a = 3, b = -N , c= 48 . 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 N = 24 or N = -24 .

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

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Example Question #2122 : Sat Mathematics

0.7 x + 0.4 y = 0.07

0.4 x+ 0.7y = -0.95

Evaluate .

Possible Answers:

The system has no solution.

x = 1 3

x = 21

x = 1.3

x = 2.1

Correct answer:

x = 1.3

Explanation:

Multiply both sides of the top equation by 7:

0.7 x + 0.4 y = 0.07

7(0.7 x + 0.4 y )= 7(0.07)

7(0.7 x )+7( 0.4 y )= 7(0.07)

4.9 x+2.8 y = 0.49

Multiply both sides of the bottom equation by :

0.4 x+ 0.7y = -0.95

-4 (0.4 x+ 0.7y )=-4 ( -0.95)

-4(0.4 x)+ (-4)(0.7y )=-4( -0.95)

-1.6x-2.8y = 3.8

Add both sides of the equations to eliminate the terms:

4.9 x+2.8 y = 0.49

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

3.3x = 4.29

Solve for :

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

x = 1.3

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Example Question #41 : Quadratic Equations

Find the solutions of the equation 2x^2-6x-36 .

Possible Answers:

x=3,x=-6

x=-3, x=-6

x=6,x=-6

x=3,x=6

x=-3, x=6

Correct answer:

x=-3, x=6

Explanation:

This is a quadratic equation because the leading term is of degree ; hence, its solutions can be easily calculated via the quadratic formula:

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

In order to use the quadratic formula to solve a quadratic equation, you must identify the values of the coefficients , , and , substitute them into the quadratic formula, and perform the arithmetical calculations to yield one, two, or no real number solutions for .

In this case, a=2 , b=-6 , and c=-36 . Hence, the quadratic formula yields

$x=\frac{-(-6)\pm\sqrt{(-6)^2-4(2)(-36)}}{2(2)}$

$=\frac{6\pm\sqrt{36+288}}{4}$

$=\frac{6\pm\sqrt{324}}{4}$

$=\frac{6\pm18}{4}$

x=6, x=-3

Hence, this equation has two real solutions: and .

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Example Question #1 : New Sat Math No Calculator

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

3x^2-5x=2x+6

Possible Answers:

$\frac{7}{3}$

$\frac{14}{15}$

$\frac{2}{3}$

Correct answer:

$\frac{7}{3}$

Explanation:

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

Ax^2+Bx+C=0

Therefore, take our equation:

3x^2-5x=2x+6

And rewrite it as:

3x^2-7x-6=0

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:

(3x+2)(x-3)=0

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 :

3x+2=0

3x=-2

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

x-3=0

x=3

The sum of these values is:

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

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Example Question #341 : Algebra

Tommy throws a rock off a 10 meter ledge at a speed of 3 meters/second. Calculate when the rock hits the ground.

To solve use the equation

s(t)=-gt^2+v_0(t)+h_0

where

$\\g=4.9 \\v_0=\text{speed} \\h_0=\text{starting height}$

Possible Answers:

$1.3\text{ seconds}$

$1.5\text{ seconds}$

$1.75\text{ seconds}$

$1.25\text{ seconds}$

$2\text{ seconds}$

Correct answer:

$1.75\text{ seconds}$

Explanation:

Tommy throws a rock off a 10 meter ledge at a speed of 3 meters/second. To calculate when the rock hits the ground first identify what is known.

Using the equation

s(t)=-gt^2+v_0(t)+h_0

where

$\\g=4.9 \\v_0=\text{speed} \\h_0=\text{starting height}$

it is known that,

$v_0=3\frac{ \text{ m}}{\text{ s}}$

$h_0=10 \text{ m}$

Substituting the given values into the position s(t) equation looks as follows.

s(t)=-4.9t^2+3(t)+10

Now to calculate when the rock hits the ground, find the value that results in s(t)=0 .

0=-4.9t^2+3t+10

Use graphing technology to graph s(t) .

Screen shot 2016 02 11 at 8.18.52 am

It appears that the rock hits the ground approximately 1.75 seconds after Tommy throws it.

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Example Question #41 : Quadratic Equations

Find the solutions for

f(x)=x^2+qx+r

Possible Answers:

$x=\frac{-q\pm\sqrt{q^2-4 q}}{2}$

$x=\frac{-q\pm\sqrt{q^2-4 r}}{2}$

$x=\frac{-r\pm\sqrt{r^2-4 q}}{2}$

$x=\frac{q\pm\sqrt{q^2-4 r}}{2}$

Correct answer:

$x=\frac{-q\pm\sqrt{q^2-4 r}}{2}$

Explanation:

The first step is to set it equal to zero.

0=x^2+qx+r

Now we will use the quadratic formula.

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

In this case a=1 , b=q , c=r

$x=\frac{-q\pm\sqrt{q^2-4\cdot 1\cdot r}}{2\cdot 1}$

$x=\frac{-q\pm\sqrt{q^2-4 r}}{2}$

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

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #31 : How To Find The Solution To A Quadratic Equation

Example Question #32 : How To Find The Solution To A Quadratic Equation

Example Question #2121 : Sat Mathematics

Example Question #2122 : Sat Mathematics

Example Question #41 : Quadratic Equations

Example Question #1 : New Sat Math No Calculator

Example Question #341 : Algebra

Example Question #41 : Quadratic Equations

All SAT Math Resources

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