Math / Algebra

Question2. Find all real solutions of the equation: $x^{2}-10 x+21=0$ . A. 1 and 21 B. -5 and -5 C. 3 and 7 $(x-7)(x-3)=0$

Studdy Solution

STEP 1

What is this asking? We need to find the values of $x$ that make the equation $x^2 - 10x + 21 = 0$ true. Watch out! There might be more than one solution, so don't stop after finding just one!
Also, make sure your solutions actually work when you plug them back into the original equation.

STEP 2

1. Factor the quadratic
2. Solve for x

STEP 3

We're dealing with a quadratic equation here, which means it has an $x^2$ term.
A common way to solve these is by factoring.
We want to rewrite our equation in the form $(x - a) \cdot (x - b) = 0$ , where $a$ and $b$ are numbers we need to figure out!

STEP 4

We need two numbers that add up to $-10$ (the coefficient of our $x$ term) and multiply to $21$ (the constant term).
Think think think... How about $-3$ and $-7$ ?
Let's check: $(-3) + (-7) = -10$ and $(-3) \cdot (-7) = 21$ .
Perfect!

STEP 5

So, we can rewrite our original equation as $(x - 3)(x - 7) = 0$ .
Look at that beautiful factoring!

STEP 6

Now, here's a cool trick called the zero product property: if two things multiplied together equal zero, then at least one of those things must be zero.
So, either $(x - 3) = 0$ or $(x - 7) = 0$ .

STEP 7

If $x - 3 = 0$ , then we can add $3$ to both sides of the equation to isolate $x$ .
This gives us $x - 3 + 3 = 0 + 3$ , which simplifies to $x = 3$ .
Boom! Our first solution is $x = 3$ .

STEP 8

If $x - 7 = 0$ , we do the same thing. Add $7$ to both sides: $x - 7 + 7 = 0 + 7$ , which simplifies to $x = 7$ .
Awesome! Our second solution is $x = 7$ .

STEP 9

The solutions to the equation $x^2 - 10x + 21 = 0$ are $x = 3$ and $x = 7$ .
So the answer is C!

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Question2. Find all real solutions of the equation: $x^{2}-10 x+21=0$ . A. 1 and 21 B. -5 and -5 C. 3 and 7 $(x-7)(x-3)=0$

Studdy Solution

STEP 1

What is this asking? We need to find the values of $x$ that make the equation $x^2 - 10x + 21 = 0$ true. Watch out! There might be more than one solution, so don't stop after finding just one!
Also, make sure your solutions actually work when you plug them back into the original equation.

STEP 2

1. Factor the quadratic
2. Solve for x

STEP 3

We're dealing with a quadratic equation here, which means it has an $x^2$ term.
A common way to solve these is by factoring.
We want to rewrite our equation in the form $(x - a) \cdot (x - b) = 0$ , where $a$ and $b$ are numbers we need to figure out!

STEP 4

We need two numbers that add up to $-10$ (the coefficient of our $x$ term) and multiply to $21$ (the constant term).
Think think think... How about $-3$ and $-7$ ?
Let's check: $(-3) + (-7) = -10$ and $(-3) \cdot (-7) = 21$ .
Perfect!

STEP 5

So, we can rewrite our original equation as $(x - 3)(x - 7) = 0$ .
Look at that beautiful factoring!

STEP 6

Now, here's a cool trick called the zero product property: if two things multiplied together equal zero, then at least one of those things must be zero.
So, either $(x - 3) = 0$ or $(x - 7) = 0$ .

STEP 7

If $x - 3 = 0$ , then we can add $3$ to both sides of the equation to isolate $x$ .
This gives us $x - 3 + 3 = 0 + 3$ , which simplifies to $x = 3$ .
Boom! Our first solution is $x = 3$ .

STEP 8

If $x - 7 = 0$ , we do the same thing. Add $7$ to both sides: $x - 7 + 7 = 0 + 7$ , which simplifies to $x = 7$ .
Awesome! Our second solution is $x = 7$ .

STEP 9

The solutions to the equation $x^2 - 10x + 21 = 0$ are $x = 3$ and $x = 7$ .
So the answer is C!

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Question2. Find all real solutions of the equation: x2−10x+21=0x^{2}-10 x+21=0x2−10x+21=0. A. 1 and 21 B. -5 and -5 C. 3 and 7 (x−7)(x−3)=0(x-7)(x-3)=0(x−7)(x−3)=0

Studdy Solution

STEP 1

STEP 2

1. Factor the quadratic2. Solve for x

STEP 3

We're dealing with a **quadratic equation** here, which means it has an x2x^2x2 term.A common way to solve these is by **factoring**.We want to rewrite our equation in the form (x−a)⋅(x−b)=0(x - a) \cdot (x - b) = 0(x−a)⋅(x−b)=0, where aaa and bbb are numbers we need to figure out!

STEP 4

STEP 5

So, we can rewrite our original equation as (x−3)(x−7)=0(x - 3)(x - 7) = 0(x−3)(x−7)=0.Look at that beautiful factoring!

STEP 6

Now, here's a cool trick called the **zero product property**: if two things multiplied together equal zero, then at least one of those things *must* be zero.So, either (x−3)=0(x - 3) = 0(x−3)=0 or (x−7)=0(x - 7) = 0(x−7)=0.

STEP 7

If x−3=0x - 3 = 0x−3=0, then we can **add** 333 to both sides of the equation to **isolate** xxx.This gives us x−3+3=0+3x - 3 + 3 = 0 + 3x−3+3=0+3, which simplifies to x=3x = 3x=3.Boom! Our **first solution** is x=3x = 3x=3.

STEP 8

If x−7=0x - 7 = 0x−7=0, we do the same thing. **Add** 777 to both sides: x−7+7=0+7x - 7 + 7 = 0 + 7x−7+7=0+7, which simplifies to x=7x = 7x=7.Awesome! Our **second solution** is x=7x = 7x=7.

STEP 9

The solutions to the equation x2−10x+21=0x^2 - 10x + 21 = 0x2−10x+21=0 are x=3x = 3x=3 and x=7x = 7x=7.So the answer is C!

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Question2. Find all real solutions of the equation: $x^{2}-10 x+21=0$ . A. 1 and 21 B. -5 and -5 C. 3 and 7 $(x-7)(x-3)=0$

1. Factor the quadratic
2. Solve for x

We're dealing with a quadratic equation here, which means it has an $x^2$ term.
A common way to solve these is by factoring.
We want to rewrite our equation in the form $(x - a) \cdot (x - b) = 0$ , where $a$ and $b$ are numbers we need to figure out!

So, we can rewrite our original equation as $(x - 3)(x - 7) = 0$ .
Look at that beautiful factoring!

Now, here's a cool trick called the zero product property: if two things multiplied together equal zero, then at least one of those things must be zero.
So, either $(x - 3) = 0$ or $(x - 7) = 0$ .

If $x - 3 = 0$ , then we can add $3$ to both sides of the equation to isolate $x$ .
This gives us $x - 3 + 3 = 0 + 3$ , which simplifies to $x = 3$ .
Boom! Our first solution is $x = 3$ .

If $x - 7 = 0$ , we do the same thing. Add $7$ to both sides: $x - 7 + 7 = 0 + 7$ , which simplifies to $x = 7$ .
Awesome! Our second solution is $x = 7$ .

The solutions to the equation $x^2 - 10x + 21 = 0$ are $x = 3$ and $x = 7$ .
So the answer is C!