Question

if alpha and beta are the roots of equation 2 x square - 35 X + 4 is equals to zero then find the value of Two Alpha minus 35 whole cube into 2 Beta -35 whole cube ​

Answer 1

Appropriate Question:

If α and β are the roots of the equation 2x² - 35x + 4 = 0, then find the value of (2α - 35)³ × (2β - 35)³

Answer:

(2α - 35)³ × (2β - 35)³ = 512

Explanation:

Given :

α and β are the roots of the equation 2x² - 35x + 4 = 0

To find :

the value of (2α - 35)³ × (2β - 35)³

Solution :

Let p(x) = 2x² - 35x + 4

For a quadratic equation of the form ax² + bx + c = 0 ;

sum of roots = -b/a

product of roots = c/a

Similarly, for the equation 2x² - 35x + 4 = 0

α + β = -(-35)/2 = 35/2

αβ = 4/2 = 2 -- eqn.[1]

Since α is a roots of the given equation, when we substitute x = α the result is zero.

p(α) = 0

2x² - 35x + 4 = 0

2(α)² - 35(α) + 4 = 0

2α² - 35α + 4 = 0

2α² - 35α = -4

α(2α - 35) = -4

 (2α - 35) = -4/α

Similarly, when we substitute x = β, the result is zero.

p(β) = 0

2x² - 35x + 4 = 0

2(β)² - 35(β) + 4 = 0

2β² - 35β + 4 = 0

2β² - 35β = -4

β(2β - 35) = -4

 (2β - 35) = -4/β

Now, we have to find the value of (2α - 35)³ × (2β - 35)³

= (2α - 35)³ × (2β - 35)³

= (-4/α)³ × (-4/β)³

$\sf = \dfrac{-64}{\alpha^3} \times \dfrac{-64}{\beta^3} \\\\ \sf = \dfrac{(-64)(-64)}{(\alpha \beta)^3} \\\\ \sf =\dfrac{4096}{2^3} \ \because eqn.[1] \\\\ \sf =\dfrac{4096}{8} \\\\ \sf =512$

Therefore, (2α - 35)³ × (2β - 35)³ = 512

Answer 2

Answer

=(2α - 35)³ × (2β - 35)³ = 512