Quality Learning Support For Better Outcomes
First time here? Checkout the FAQs!
x
MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

0 like 0 dislike
15 views
If the roots of the quadratic equation $x^{2}+p x+q=0$ are $\tan 30^{\circ}$ and $\tan 15^{\circ}$ respectively then the value of $2+q-p$ is

1) 2
2) 3
3) 0
4) 1
in Mathematics by Bronze Status (9,486 points) | 15 views

1 Answer

0 like 0 dislike
Best answer

Answer

(2)

Explanation

$x^{2}+p x+q=0$


$\tan 30^{\circ}+\tan 15^{0}=-p$


$\tan 30^{\circ} \cdot \tan 15^{0}=q$


$\tan 45^{\circ}=\frac{\tan 30^{\circ}+\tan 15^{\circ}}{1-\tan 30^{0} \tan 15^{\circ}}=\frac{-p}{1-q}=1$


$\Rightarrow-p=1-q$


$\Rightarrow q-p=1 \quad \therefore 2+q-p=3$

by Bronze Status (9,486 points)

Join the MathsGee community and get study support for success - MathsGee provides answers to subject-specific educational questions for improved outcomes.



On MathsGee Answers, you can:


1. Ask questions
2. Answer questions
3. Comment on Answers
4. Vote on Questions and Answers
5. Donate to your favourite users

Enter your email address:

MathsGee Tools

Math Worksheet Generator

Math Algebra Solver

Trigonometry Simulations

Vectors Simulations

Matrix Arithmetic Simulations

Matrix Transformations Simulations

Quadratic Equations Simulations

Probability & Statistics Simulations

PHET Simulations

Visual Statistics

MathsGee ZOOM | eBook