If $x=\sqrt{2}$ is one of the roots of the equation $ax^{2}+\sqrt{2}bx+2c=0$; $a \neq 0$,$a, b, c \in R$,then prove that $a+b+c=0$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Given that $x=\sqrt{2}$ is one of the roots of the quadratic equation $ax^{2}+\sqrt{2}bx+2c=0$.
Since $x=\sqrt{2}$ is a root,it must satisfy the equation.
Substituting $x=\sqrt{2}$ into the equation:
$a(\sqrt{2})^{2} + \sqrt{2}b(\sqrt{2}) + 2c = 0$
Simplifying the terms:
$a(2) + 2b + 2c = 0$
$2a + 2b + 2c = 0$
Dividing the entire equation by $2$:
$a + b + c = 0$
Hence,it is proved that $a+b+c=0$.

Explore More

Similar Questions

The product of the digits of a two-digit number is $14$. The number obtained by interchanging the digits is $45$ more than the original number. Find the original number.

Difficult
View Solution

Find whether the following equation has real roots. If real roots exist,find them.
$5x^{2}-2x-10=0$

Obtain the roots of the following quadratic equation using the quadratic formula: $\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0$.

The roots of the quadratic equation $x^{2}-2x-15=0$ are ....... .

When there is a decrease of $10 \,km/hr$ in the usual speed of a train,it takes $4 \frac{1}{2}$ hours more to cover a $900 \,km$ distance. Find the usual speed of the train.

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo