JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

\mathrm{p}_1

and

\mathrm{p}_2

are the lengths of the perpendiculars from the points

(2,3,4)

and

(1,1,4)

respectively on the plane

3 x-6 y+2 z+11=0

, then

p_1, p_2

are the roots of the equation

View answer and explanation

Correct option: B

7 \mathrm{p}^2-23 \mathrm{p}+16=0

Option B (

7 p^2-23 p+16=0

) is correct because it involves calculating the perpendicular distances from the given points to the plane and then forming a quadratic equation with those distances as roots.

More Questions from QUADRATIC EQUATIONS

Practice All Chapter MCQs →

The solution of the equation

\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0

is If

\alpha ,\beta

are the roots of

{x^2} - 3x + a = 0,a \in R

and

\alpha < 1 < \beta ,

then If

\alpha ,\;\beta

be the roots of the equation

{x^2} - px + q = 0

and

{\alpha _1},{\beta _1}

be the roots of the equation

{x^2} - qx + p = 0,

then the equation whose roots are

\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}},

is If

{x^2} + 4ax + 2 > 0

for all values of

x

, then

a

lies in the interval If

f\left( x \right) = a{x^2} + bx + c\;

and

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

and

\;a \ne 0,

then Root(s) of the equation

9{x^2} - 18\left| x \right| + 5 = 0

belonging to the domain of definition of the function

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)

is (are)