Question

If a and b are the zeroes of of
p(x)=4x2-6x+2 then find a2-b2

Answer 1

ATQ, "a" and "b" are the zeroes of polynomial 4x² - 6x + 2

let's find it's zeroes first using splitting method.

= 4x² - 6x + 2

= 4x² - (4x + 2x) + 2

= 4x² - 4x - 2x + 2

= 4x(x - 1) - 2(x - 1)

= (x - 1) (4x - 2)

equating both factors by 0

• x - 1 = 0

➡ x = 1

• 4x - 2 = 0

➡ x = 1/2

a = 1 and b = 1/2

hence, value of a² - b²

= (1)² - (1/2)²

= 1 - 1/4

= (4 - 1)/4

= 3/4

Answer 2

Answer:

$\large{\underline{\boxed{\sf \dfrac{3}{4}} }}$

Step-by-step explanation:

Given that;

a and b are the zeroes of the equation 4x² - 6x + 2.

To find the value of a and b, we need to factorise the given equation. It can be solved by splitting the middle term ;

4x² - 6x + 2

⇒ 4x² - (4 + 2)x + 2

⇒ 4x² - 4x - 2x + 2

⇒ 4x(x - 1) - 2(x - 1)

⇒ (x - 1)(4x - 2)

By zero product rule ;

⇒ (x - 1) = 0 or (4x - 2) = 0

⇒ x = 1 or x = $\sf \dfrac{2}{4} = \dfrac{1}{2}$

Now,

a² - b²

$\sf \implies (1) {}^{2} - ( \frac{1}{2} ) {}^{2} \\ \\ \sf \implies 1 - \frac{1}{4} \\ \\ \sf \implies \frac{4 - 1}{4} \\ \\ \sf \implies \frac{3}{4}$

Hence, the answer is 3/4.