Question

find the value of p so that each pair of the equation may have one root common.(1) 2x^2+px-1=0 and 3x^2-2x-5=0​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm \: \alpha ,\gamma \: be \: the \: roots \: of {3x}^{2} - 2x - 5 = 0$

Consider,

$\rm \: {3x}^{2} - 2x - 5 = 0$

$\rm \: {3x}^{2} - 5x + 3x - 5 = 0$

$\rm \: x(3x - 5) + 1(3x - 5) = 0$

$\rm \: (x + 1)(3x - 5) = 0$

$\rm\implies \:x = - 1 \: \: or \: \: x = \dfrac{5}{3}$

$\rm\implies \: \gamma = - 1 \: \: or \: \: \alpha = \dfrac{5}{3} \\ \rm \: or \\ \rm\implies \: \alpha = - 1 \: \: or \: \: \gamma = \dfrac{5}{3}$

Now,

Let assume that,

$\rm \: \alpha, \beta \: be \: the \: roots \: of \: {2x}^{2} + px - 1 = 0$

We know,

$\boxed{\red{\sf Sum\ of\ the\ roots=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm\implies \: \alpha + \beta = - \dfrac{p}{2}$

Also,

$\boxed{\red{\sf Product\ of\ the\ roots=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm\implies \: \alpha \beta = - \dfrac{1}{2}$

It means, we have

$\rm\implies \: \alpha + \beta = - \dfrac{p}{2} \: \: and \: \: \alpha \beta = - \dfrac{1}{2}$

Case :- 1

$\rm \: \alpha = - 1$

As, we have

$\rm \: \alpha \beta = - \dfrac{1}{2}$

$\rm \: ( - 1) \beta = - \dfrac{1}{2}$

$\rm \:\beta = \dfrac{1}{2}$

Now,

$\rm \: \alpha + \beta = - \dfrac{p}{2}$

$\rm \: - 1 + \dfrac{1}{2} = - \dfrac{p}{2}$

$\rm \: - \dfrac{1}{2} = - \dfrac{p}{2}$

$\rm\implies \:p \: = \: 1$

Now, Consider Case :- 2

$\rm \: \alpha = \dfrac{5}{3}$

As, we have

$\rm \: \alpha \beta = - \dfrac{1}{2}$

$\rm \: \frac{5}{3} \times \beta = - \dfrac{1}{2}$

$\rm \: \beta = - \dfrac{3}{10}$

Now,

$\rm \: \alpha + \beta = - \dfrac{p}{2}$

$\rm \: \dfrac{5}{3} - \dfrac{3}{10} = - \dfrac{p}{2}$

$\rm \: \dfrac{50 - 9}{30} = - \dfrac{p}{2}$

$\rm \: \dfrac{41}{30} = - \dfrac{p}{2}$

$\rm\implies \:p \: = \: - \: \dfrac{41}{15}$

So,

$\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\bf{p \: = \: 1} \\ \\ &\sf{or}\\ \\ &\bf{p \: = \: - \: \dfrac{41}{15} } \end{cases}\end{gathered}\end{gathered}$

$\rule{190pt}{2pt}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

Answer 2

Answer:

Given:-

The two equations are:-

• 2x² + px - 1 = 0
• 3x² - 2x - 5 = 0

It is also given that the two equations may have one root common.

So, first, we have to find the roots of second equation as:

3x²-2x-5=0

»» 3x²+3x-5x-5=0

»» 3x×(x+1)-5×(x+1)=0

»» (x+1) × (3x-5) =0

So, the roots of x for the second equation are,

• x = -1

or

• x = 5/3

If we substitute these values of x in the first equation, we can get the possible values of p.

2x²+px-1=0

»» 2×(-1)²+p×(-1)-1=0

»» 2-p-1 = 0

»» p = 1

Or

»» 2×(5/3)² + p×5/3-1 = 0

»» 2×25/9+5p/3=1

»» 50+15p = 9

»» 15p = 9-50

»» p = -41/15

Therefore, the possible values of p are 1 or -41/15 , So that the given two equations may have a common root.