......... can be taken as the value of k,if o | vedclass

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

Question

(A) Given the quadratic equation $k x^{2} - 7x + 6 = 0$.Since $\frac{3}{2}$ is a root of the equation,it must satisfy the equation.Substitute $x = \fr

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given the quadratic equation $k x^{2} - 7x + 6 = 0$.Since $\frac{3}{2}$ is a root of the equation,it must satisfy the equation.Substitute $x = \frac{3}{2}$ into the equation:$k(\frac{3}{2})^{2} - 7(\frac{3}{2}) + 6 = 0$$k(\frac{9}{4}) - \frac{21}{2} + 6 = 0$Multiply the entire equation by $4$ to clear the denominators:$9k - 42 + 24 = 0$$9k - 18 = 0$$9k = 18$$k = 2$Therefore,the value of $k$ is $2$.

......... can be taken as the value of $k$,if one of the roots of the quadratic equation $k x^{2} - 7x + 6 = 0$ is $\frac{3}{2}$.

Similar Questions

Solve the following equation using the method of factorization: $\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6} \,(x \neq 3, -5)$

$x = \dots$ is a solution of the quadratic equation $x^{2} + 7x + 12 = 0$.

State whether the quadratic equation $x(1-x)-2=0$ has two distinct real roots. Justify your answer.

Find the discriminant of the following quadratic equation: $3x^{2} - 12x + 16 = 0$.

Obtain the roots of the following quadratic equation by using the quadratic formula: $p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module