Obtain the roots of the following equation us | vedclass

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

Question

(B) To solve the equation $m^{2} - 18m + 81 = 0$ by the method of completing the square:$1$. The equation is already in the form $m^{2} - 18m = -81$.$

Vedclass · Accepted Answer

$9, 9$

Vedclass · Answer

(B) To solve the equation $m^{2} - 18m + 81 = 0$ by the method of completing the square:$1$. The equation is already in the form $m^{2} - 18m = -81$.$2$. To complete the square,we add the square of half the coefficient of $m$ to both sides. The coefficient of $m$ is $-18$,so half of it is $-9$,and $(-9)^{2} = 81$.$3$. Adding $81$ to both sides: $m^{2} - 18m + 81 = -81 + 81$.$4$. This simplifies to $(m - 9)^{2} = 0$.$5$. Taking the square root of both sides,we get $m - 9 = 0$,which implies $m = 9$.$6$. Since it is a quadratic equation,the roots are $9, 9$.

Obtain the roots of the following equation using the method of 'completing the square': $m^{2} - 18m + 81 = 0$

Similar Questions

Solve the following quadratic equation using the quadratic formula: $3x^{2} + 2\sqrt{5}x - 5 = 0$.

Examine whether the following equation is quadratic or not: $(x-3)(2x+1) = x(x+5)$

State whether the following statements are true or false. Justify your answers.
$(i)$ Every quadratic equation has at most two roots.
$(ii)$ If the coefficient of $x^{2}$ and the constant term of a quadratic equation have opposite signs,then the quadratic equation has real roots.

The roots of the quadratic equation $x + \frac{2}{x} = 3$ are $\dots$ $(x \neq 0)$.

$(x+6)(x+5)=0$ is a ....... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module