If 3 is a root of x^2 + kx - 24 = 0,it is als | vedclass

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

Question

(C) Given that $3$ is a root of the equation $x^2 + kx - 24 = 0$.Substituting $x = 3$ into the equation:$(3)^2 + k(3) - 24 = 0$$9 + 3k - 24 = 0$$3k -

Vedclass · Accepted Answer

$x^2 - kx + 6 = 0$

Vedclass · Answer

(C) Given that $3$ is a root of the equation $x^2 + kx - 24 = 0$.Substituting $x = 3$ into the equation:$(3)^2 + k(3) - 24 = 0$$9 + 3k - 24 = 0$$3k - 15 = 0$$3k = 15$$k = 5$Now,check which option satisfies the condition that $x = 3$ is a root when $k = 5$:For option $(C)$,$x^2 - kx + 6 = 0$ becomes $x^2 - 5x + 6 = 0$.Substituting $x = 3$: $(3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$.Thus,$3$ is a root of $x^2 - kx + 6 = 0$.

If $3$ is a root of $x^2 + kx - 24 = 0$,it is also a root of

Similar Questions

If $mn=3$ and $\frac{1}{m}+\frac{1}{n}=\frac{4}{3}$,then the value of $0.1+0.1^{\frac{1}{m}}+0.1^{\frac{1}{n}}$ is

If $(x-1)$ is a factor of the polynomial $x^{5}-4 x^{3}+2 x^{2}-3 x+k=0$,then the value of $k$ is:

If $f(x)=2 x^4-13 x^2+a x+b$ is divisible by $x^2-3 x+2$,then $(a, b)$ is equal to

If $ax^2 + bx + c = 0$ has real and distinct roots,$\alpha$ and $\beta$ where $\beta > \alpha$. Further,if $a > 0, b < 0$,and $c < 0$,then:

If $x = \sqrt{7} + \sqrt{3}$ and $xy = 4$,then $x^4 + y^4 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module