Solve (x+5)² - 49: Finding X-Axis Intersections

Q: Why do we set y = 0 to find x-intercepts?

X-intercepts are points where the graph crosses the x-axis. On the x-axis, the y-coordinate is always 0! So we solve $ (x+5)^2 - 49 = 0 $ to find where y = 0.

Q: What does the ± symbol mean when taking square roots?

The plus-minus symbol (±) means we consider both positive and negative roots. Since $ 7^2 = 49 $ AND $ (-7)^2 = 49 $, we get $ \sqrt{49} = ±7 $.

Q: How do I write the final answer as coordinate points?

X-intercepts are written as (x, 0) because the y-coordinate is always 0 on the x-axis. So our solutions x = -12 and x = 2 become the points $ (-12, 0) $ and $ (2, 0) $.

Q: Can I expand the squared term instead of using this method?

Yes, but it's much harder! You'd get $ x^2 + 10x + 25 - 49 = 0 $, then $ x^2 + 10x - 24 = 0 $ and need the quadratic formula. The square root method is faster for perfect square forms.

Q: What if the number under the square root isn't a perfect square?

You'll get irrational solutions with square root symbols. For example, if you had $ (x+5)^2 = 50 $, you'd get $ x + 5 = ±\sqrt{50} = ±5\sqrt{2} $.

Quadratic Functions with X-Intercept Solutions

Find the intersection of the function

$y=(x+5)^2-49$

With the X

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the intersection point with the X-axis

00:03 Substitute Y=0 and solve to find the intersection point

00:10 We want to isolate X

00:22 Extract the root

00:30 When extracting a root there are 2 solutions, positive and negative

00:38 Solve each possibility to find the intersection point

00:58 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the intersection of the function

$y=(x+5)^2-49$

With the X

Step-by-step solution

To find the intersection of the function $y = (x+5)^2 - 49$ with the x-axis, we need to solve the equation for $y = 0$ .

Set the equation equal to zero:

$(x+5)^2 - 49 = 0$

Add 49 to both sides:

$(x+5)^2 = 49$

Take the square root of both sides, remembering to consider both the positive and negative roots:

$x + 5 = 7$ or $x + 5 = -7$

Solve for $x$ in both cases:

For $x + 5 = 7$ :
$x = 7 - 5$
$x = 2$
For $x + 5 = -7$ :
$x = -7 - 5$
$x = -12$

Therefore, the x-intercepts of the function are $(-12, 0)$ and $(2, 0)$ .

Thus, the function intersects the x-axis at these points.

$(-12, 0), (2, 0)$

Final Answer

$(-12,0),(2,0)$

Key Points to Remember

Essential concepts to master this topic

Rule: Set function equal to zero to find x-intercepts
Technique: Take square root of both sides: $\sqrt{49} = \pm 7$
Check: Substitute x-values back to verify y = 0 for both points ✓

Common Mistakes

Avoid these frequent errors

Forgetting the negative square root
Don't only consider $x + 5 = 7$ when solving $(x+5)^2 = 49$ = missing one x-intercept! This gives only x = 2 instead of both solutions. Always remember that $\sqrt{49} = \pm 7$ , so solve both $x + 5 = 7$ AND $x + 5 = -7$ .

Practice Quiz

Test your knowledge with interactive questions

Find the corresponding algebraic representation of the drawing:

FAQ

Everything you need to know about this question

Why do we set y = 0 to find x-intercepts?

X-intercepts are points where the graph crosses the x-axis. On the x-axis, the y-coordinate is always 0! So we solve $(x+5)^2 - 49 = 0$ to find where y = 0.

What does the ± symbol mean when taking square roots?

The plus-minus symbol (±) means we consider both positive and negative roots. Since $7^2 = 49$ AND $(-7)^2 = 49$ , we get $\sqrt{49} = ±7$ .

How do I write the final answer as coordinate points?

X-intercepts are written as (x, 0) because the y-coordinate is always 0 on the x-axis. So our solutions x = -12 and x = 2 become the points $(-12, 0)$ and $(2, 0)$ .

Can I expand the squared term instead of using this method?

Yes, but it's much harder! You'd get $x^2 + 10x + 25 - 49 = 0$ , then $x^2 + 10x - 24 = 0$ and need the quadratic formula. The square root method is faster for perfect square forms.

What if the number under the square root isn't a perfect square?

You'll get irrational solutions with square root symbols. For example, if you had $(x+5)^2 = 50$ , you'd get $x + 5 = ±\sqrt{50} = ±5\sqrt{2}$ .

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