MCQ
A quadratic equation whose one root is $3$ is :
- A$ x^2-5 x-6=0 $
- B$ x^2-6 x-6=0 $
- ✓$ x^2-5 x+6=0 $
- D$ x^2+6 x-5=0 $
✓
Answer
Correct option: C.
$ x^2-5 x+6=0 $
since $3$ is the root of the equation $, x = 3$ must satisfy the equation.
Applying $x = 3$ in the equation $ x^2-5 x+6=0 $
gives, $(3)^2- 5(3) + 6 = 0$
$\Rightarrow 9 - 15 + 6 =0$
$\Rightarrow 15 - 15 = 0$
$\Rightarrow 0 = 0$
$\Rightarrow \text{L.H.S. = R.H.S}$
$ x^2-5 x+6=0 $ is a required equation which has $3$ as root.
Applying $x = 3$ in the equation $ x^2-5 x+6=0 $
gives, $(3)^2- 5(3) + 6 = 0$
$\Rightarrow 9 - 15 + 6 =0$
$\Rightarrow 15 - 15 = 0$
$\Rightarrow 0 = 0$
$\Rightarrow \text{L.H.S. = R.H.S}$
$ x^2-5 x+6=0 $ is a required equation which has $3$ as root.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If one zero of the quadratic polynomial $kx^2 + 3x + k$ is $2$, then the value of $k$ is:
→From a point $Q,$ the length of tangent to a circle is $24\ cm$ and the distance of $Q$ from the centre is $25\ cm,$ radius of circle is :
→The value of $'k\ ’$ for which the system of equations $3x + 5y = 0$ and $kx + 10y = 0$ has a non zero solution is :
→The LCM of two numbers is 1200 . Which of the following cannot be their HCF?
If the circumference of a circle and the perimeter of a square are equal, then:
If the perimeter and area of a circle are numerically equal, then the radius of the circle is :
If (3, –6) is the mid-point of the line segment joining (0, 0) and (x, y), then the point (x, y) is:
Two friends Rohit and Mohit are standing on the opposite sides of a tower of height 60 metres. If their angles of depression seen from the top of the tower are $30^{\circ}$ and $45^{\circ}$ respectively, then the distance between the two friends is
→The ratio in which the point $(4,0)$ divides the line segment joining the points $(4,6)$ and $(4,-8)$ is
→If the system of equations $2 x+3 y=5,4 x+k y=10$ has infinitely many solutions, then $k=$
→