Find the roots of the following quadratic equ | vedclass

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

Question

(A) Given equation: $2x^{2} - 7x + 3 = 0$.Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:$x^{2} - \frac{7}{2}x + \f

Vedclass · Accepted Answer

$3, \frac{1}{2}$

Vedclass · Answer

(A) Given equation: $2x^{2} - 7x + 3 = 0$.Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:$x^{2} - \frac{7}{2}x + \frac{3}{2} = 0$.Shift the constant term to the right side:$x^{2} - \frac{7}{2}x = -\frac{3}{2}$.Add the square of half the coefficient of $x$ (which is $\frac{1}{2} 	imes \frac{7}{2} = \frac{7}{4}$) to both sides:$x^{2} - \frac{7}{2}x + (\frac{7}{4})^{2} = -\frac{3}{2} + (\frac{7}{4})^{2}$.$(x - \frac{7}{4})^{2} = -\frac{3}{2} + \frac{49}{16}$.$(x - \frac{7}{4})^{2} = \frac{-24 + 49}{16} = \frac{25}{16}$.Taking the square root on both sides:$x - \frac{7}{4} = \pm \frac{5}{4}$.Case $1$: $x = \frac{7}{4} + \frac{5}{4} = \frac{12}{4} = 3$.Case $2$: $x = \frac{7}{4} - \frac{5}{4} = \frac{2}{4} = \frac{1}{2}$.Thus,the roots are $3$ and $\frac{1}{2}$.

Find the roots of the following quadratic equation by the method of completing the square: $2x^{2} - 7x + 3 = 0$.

Similar Questions

If the discriminant of $x^{2}-10x+(2k-1)=0$ is $40$,then $k=$...............

If the following quadratic equation has two equal and real roots,then find the value of $k$: $k x^{2} - 2 \sqrt{5} x + 4 = 0$.

Solve the following equation using the method of factorization: $x - \frac{20}{x} = 1$

Find the roots of the following quadratic equation by the factorisation method:
$3 \sqrt{2} x^{2}-5 x-\sqrt{2}=0$

The solution set of a quadratic equation $x^{2}+5x-14=0$ is $\ldots \ldots \ldots \ldots .$

Vedclass Test Series

Exam Paper Generator

Online Exam Module