Which constant should be added and subtracted | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To solve the quadratic equation $4x^{2} - \sqrt{3}x - 5 = 0$ by the method of completing the square,we first ensure the coefficient of $x^{2}$ is

Vedclass · Accepted Answer

$\frac{3}{16}$

Vedclass · Answer

(B) To solve the quadratic equation $4x^{2} - \sqrt{3}x - 5 = 0$ by the method of completing the square,we first ensure the coefficient of $x^{2}$ is $1$ or a perfect square. Here,$4x^{2} = (2x)^{2}$.We rewrite the equation as $(2x)^{2} - 2(2x)(\frac{\sqrt{3}}{4}) - 5 = 0$.To complete the square in the form $(a - b)^{2} = a^{2} - 2ab + b^{2}$,we identify $a = 2x$ and $b = \frac{\sqrt{3}}{4}$.The term to be added and subtracted is $b^{2} = (\frac{\sqrt{3}}{4})^{2} = \frac{3}{16}$.Thus,the constant is $\frac{3}{16}$.

Which constant should be added and subtracted to solve the quadratic equation $4x^{2} - \sqrt{3}x - 5 = 0$ by the method of completing the square?

Similar Questions

The length of a rectangular garden exceeds its breadth by $6 \, m$. If the area of the garden is $216 \, m^2$,find the length and the breadth of the garden.

Solve the following equation using the method of factorization: $\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6} \,(x \neq 1, -1)$

Solve the following equation using the method of factorization and write its solution set: $\frac{9}{x-1} - \frac{2}{x-3} = \frac{5}{x+1}$

If the following quadratic equation has two equal and real roots,then find the value of $k$: $2kx^2 - 8x + k = 0$.

If one of the roots of the equation $px^{2} + qx + r = 0$ where $p \neq 0$ and $p, q, r \in R$ is $x = -2$,then prove that $4p - 2q + r = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module