Question

If p(x) =2x²-3x+7 then find the value of alpha ²-beta²

Answer 1

$\huge \pink{ \mathfrak{soultion}}$

Given :

p(x)=

$2 {x}^{2} - 3x + 7$

Here,

a= 2

b=-3

c=7

As we know that

$\alpha + \beta = \frac{ - b}{a} = \frac{ - ( - 3)}{2} = \frac{3}{2}$

and

$\alpha \beta = \frac{c}{a} = \frac{7}{2}$

To find :

$\leadsto { \alpha }^{2} - { \beta }^{2} \\ \\ \rightarrow \: ( \alpha + \beta )( \alpha - \beta ) \\ \\ \rightarrow( \alpha + \beta )( \sqrt{( \alpha + \beta )^{2} - 4 \alpha \beta } ) \\ \\ \rightarrow \: ( \frac{3}{2} )( \sqrt{ (\frac{3}{2})^{2} - 4 \times \frac{7}{2} } )\\ \\ \rightarrow( \frac{3}{2} )( \sqrt{ \frac{9}{4} - 14 } ) \\ \\ \rightarrow \: \frac{3}{2} \sqrt{ \frac{ - 47}{4} } \\ \\ \rightarrow \: \frac{3}{2} ( \frac{47i}{2} ) \\ \\ \rightarrow \: \frac{3 \times 47i}{4}$

●√-1=i

Answer 2

SOLUTION:-

Given:

If p(x)=2x²-3x+7.

To find:

The value of α² -β².

Explanation:

We know that formula of the α² - β²,

⇒ (α+β)(α-β)=(α+β)√(α+β)²-4αβ

&

We can compare with the equation Ax² +Bx+C.

• A= 2
• B= -3
• C= 7

Here,

• Sum of the zeroes:

⇒ α+β= $\frac{-b}{a}$

⇒ α+β= $\frac{-(-3)}{2}$

⇒ α+β= $\frac{3}{2}$

• Product of the zeroes:

⇒ αβ= $\frac{c}{a}$

⇒ αβ= $\frac{7}{2}$

Now, putting the value of the above formula:

alpha² - beta² = (α+β)√(α+β)²-4αβ

$\frac{3}{2} \sqrt{(\frac{3}{2} )^{2} -4* \frac{7}{2} } \\\\\frac{3}{2} \sqrt{\frac{9}{4} -\frac{28}{2} }$

$\frac{3}{2} \sqrt{\frac{9-56}{4} } \\\\\frac{3}{2} \sqrt{\frac{-47}{4} } \\\frac{3}{2} *\frac{\sqrt{-47} }{2} \\\frac{3*\sqrt{-47} }{4}$.

Thus,

3×47i/4           [i=√-1]

Remark;

The formal definition of i is i^2=-1 and not √-1=i,