2 years ago

What is the root of the quadratic equation y= (x-4) (x+7)

Mathematics

1 answer:

pishuonlain [190]2 years ago

4 0

Root are also known as what x is equal to. So, to find x you would need to set (x-4)=0 and (x+7)=0
X=4. X=-7

You might be interested in

Solve the equation in the complex number system <br> X^4-2x^2-3=0

Anastaziya [24]

$x^4-2x^2-3=0\\
x^4+x^2-3x^2-3=0\\
x^2(x^2+1)-3(x^2+1)=0\\
(x^2-3)(x^2+1)=0\\
x^2-3=0 \vee x^2+1=0\\
x^2=3 \vee x^2=-1\\
x=-\sqrt3 \vee x=\sqrt3 \vee x=-i \vee x=i$

5 0

3 years ago

Use Pythagorean Theorem to solve for D. Round your answer to the nearest tenth

Aneli [31]

The answer is B. 18.9

3 0

2 years ago

Read 2 more answers

Ravi bought a desk on sale for $544. This price was 15% less than the original price. What was the original price?

mart [117]

The original price is $625.60. Hope this helps !!

8 0

3 years ago

Read 2 more answers

Question is timed! Please Answer ASAP

Lisa [10]

Answer:

The answer is A.

Step-by-step explanation:

the 1/3 compresses it, and the (-) reflects it over the y axis.

4 0

2 years ago

vagabundo [1.1K]

$\left(\sqrt x - \sqrt{3y} \right) \left(\sqrt x + \sqrt{3y} \right)\\\\=\left(\sqrt x\right)^2 - \left(\sqrt{3y} \right)^2\\\\=x -3y$

7 0

2 years ago

Read 2 more answers