Question

if the zeros of p(x)=3x^2-x-4 are the zeros of alpha and bitta then alpha square bitta +alpha bitta square= a)4/9 b)-4 c)-4/9 d)4​

Answer 1

Question : -

If the zeroes of p ( x ) = 3x² - x - 4 are alpha ( α ) & beta ( β ) . Then find the value of α²β + αβ² ?

• a ) 4/9

• b ) - 4

• c ) - 4/9

• d ) 4

Answer : -

Given : -

Quadratic equation

p ( x ) = 3x² - x - 4

Required to find : -

• value of α²β + αβ²

Solution : -

p ( x ) = 3x² - x - 4

The standard form of a quadratic equation is ax² + bx + c = 0

Let's compare this standard form with given quadratic equation .

Here,

• a = 3

• b = - 1

• c = - 4

Recall the relationship between the zeroes of the polynomial and the coefficients .

So,

We know that ;

The relationship between the sum of the zeroes and the coefficients is ;

α + β = - coefficient of x/ coefficient of x²

α + β = - b/a

α + β = - ( - 1 )/3

α + β = 1/3

Hence,

• α + β = 1/3

Similarly,

The relationship between the product of the zeroes

α.β = constant term/ coefficient of x²

α.β = c/a

α.β = - 4/3

Hence,

• α.β = - 4/3

Now,

Let's find the value of α²β + αβ² !

α²β + αβ²

Taking α.β common

αβ( α + β )

Since,

• α.β = - 4/3

• α + β = 1/3

This implies ;

- 4/3 ( 1/3 )

- 4/9

Hence,

• α²β + αβ² = - 4/9

Option - c is correct ✓

Additional Information : -

The quadratic formula is ;

$\boxed{\tt{ x = \dfrac{ - b \pm \sqrt{ b^2 - 4ac }}{ 2a} }}$

This formula helps us to find the roots of the equation .

However,

This formula can also splited and written as.

$\tt{ \alpha = \dfrac{-b + \sqrt{b^2 - 4ac }}{ 2a } }$

$\tt{ \beta = \dfrac{ - b - \sqrt{ b^2 - 4ac }}{ 2a } }$

Here,

α , β are the roots of the quadratic equation .

Now,

Let's discuss about an interesting trick .

Here, in the quadratic formula . b² - 4ac is a discriminate .

This is because it can tells about the nature of the roots .

Discriminate is represented by letter "D".

The conditions are as follows ;

• If D = 0

The roots are equal and real .

• If D > 0

The roots are unequal and rational ( if it is a perfect square )

• If D > 0

The roots are distinct and irrational ( if it is not a perfect square )

• If D < 0

The roots are unequal and imaginary .

Answer 2

GIVEN :-

• p(x) = 3x² - x - 4 with α and β as zeroes

TO FIND :-

• α²β + β²α

SOLUTION :-

Given equation is 3x² - x - 4 = 0

$\large\tt{\rightarrow{3x^2+3x-4x-4\:=\:0}}$

$\large\tt{\rightarrow{3x(x+1)-4(x+1)}}$

$\large\tt{\rightarrow{(3x-4)(x+1)\:=\:0}}$

$\large\tt{\rightarrow{x\:=\:\frac{4}{3}\:(or)\:x\:=\:-1}}$

Given Roots are α , β so,

• α = 4/3
• β = -1

Or ,

• α = -1
• β = 4/3

Consider α = 4/3 , β = -1

Then,

$\alpha {}^{2} \beta + \beta {}^{2} \alpha = ( \frac{4}{3} ) {}^{2} .( - 1) + ( - 1) {}^{2} .( \frac{4}{3} ) \\ \\ - > \alpha {}^{2} \beta + \beta {}^{2} \alpha = - \frac{16}{9} + ( \frac{4}{3} ) \\ \\ - > \alpha {}^{2} \beta + \beta {}^{2} \alpha = \frac{ - 16 + 12}{9} \\ \\ - > \alpha {}^{2} \beta + \beta {}^{2} \alpha = \frac{ - 4}{9}$

Consider α = -1 , β = 4/3

$\alpha {}^{2} \beta + \beta {}^{2} \alpha = ( -1) {}^{2} .( \frac{4}{3} ) + ( \frac{4}{3} ) {}^{2} .( - 1) \\ \\ - > { \alpha }^{2} \beta + { \beta }^{2} \alpha = \frac{4}{3} + ( - \frac{16}{9} ) \\ \\ - > { \alpha }^{2} \beta + { \beta }^{2} \alpha = \frac{ - 16 + 12}{9} \\ \\ - > { \alpha }^{2} \beta + { \beta }^{2} \alpha = \frac{ - 4}{9}$

In both cases Value is same So answer is -4/9

∴ α²β + β²α = -4/9

Hence , Option(C) is correct

$\large\tt{\underline{\underline{\green{Additional\:infirmation:-}}}}$

❃ In a quadratic equation ax² + bx + c = 0 , x is given by ,

$\boxed{\rm{ x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{ 2a }}}$