Question

find zeroes and relation between zeroes and coficience of polynomial 2x²+X-3

Answer 1

Answer:

Given :-

• 2x² - x - 3

To Find :-

What are the zeros of the quadratic polynomial and verify the relationship between the zeroes and co-efficient.

$\pink{ \sf \bold{Solution :-}}$

Given Equation :

$\begin{gathered} \boxed{\bigstar \: \: \sf\bold{\purple {2x^2 - x - 3}}}\\\end{gathered}$

$\mapsto \sf 2x^2 - x - 3 =\: 0$

$\implies \sf 2x^2 - (3 - 2)x - 3 =\: 0$

$\implies \sf 2x^2 - 3x + 2x - 3 =\: 0$

$\implies \sf x(2x - 3) + 1(2x - 3) =\: 0$

$\implies \bf (x + 1) =\: 0$

$\longrightarrow \sf x + 1 =\: 0$

$\longrightarrow \sf\bold{\red{x =\: - 1}}$

$\implies \bf{(2x - 3) =\: 0}$

$\longrightarrow \sf 2x =\: 3$

$\begin{gathered}\longrightarrow \sf\bold{\red{x =\: \dfrac{3}{2}}}\\\end{gathered}$

$\begin{gathered} \large{{\small{\bold{\underline{\therefore\: \red{ The\: zeroes\: are\: - 1\: and\: \dfrac{3}{2}\: respectively\: .}}}}}} \\ \end{gathered}$

Now, we have to verify the relationship between the zeroes and co-efficient ;

Given Equation :

$\bigstar\: \bf{2x^2 - x - 3}$

where,

• a = 2
• b = - 1
• c = - 3

➲ Sum of Roots :

As we know that :

$\begin{gathered}\mapsto \sf\boxed{\bold{\pink{Sum\: of\: roots\: (\alpha + \beta) =\: \dfrac{- b}{a}}}}\\\end{gathered}$

We have :

• a = 2
• b = - 1
• α = - 1
• β = 3/2

According to the question by using the formula we get,

$\implies \sf (- 1) + \dfrac{3}{2} =\: \dfrac{- (- 1)}{2}$

$\implies \sf - 1 + \dfrac{3}{2} =\: \dfrac{1}{2}$

$\implies \sf \dfrac{- 2 + 3}{2} =\: \dfrac{1}{2}$

$\begin{gathered}\implies \sf\bold{\green{\dfrac{1}{2} =\: \dfrac{1}{2}}}\\\end{gathered}$

Hence, Verified.

➲ Product of Roots :

As we know that :

$\begin{gathered}\mapsto \sf\boxed{\bold{\pink{Product\: of\: roots\: (\alpha\beta) =\: \dfrac{c}{a}}}}\\\end{gathered}$

We have :

• a = 2
• c = - 3
• α = - 1
• β = 3/2

According to the question by using the formula we get,

$\implies \sf (- 1) \times \dfrac{3}{2} =\: \dfrac{- 3}{2}$

$\sf \: \implies \sf - 1 \times \dfrac{3}{2} =\: \dfrac{3}{2}$

$\implies \sf \bold{\green{\dfrac{- 3}{2} =\: \dfrac{- 3}{2}}}$

Hence, Verified.

Answer 2

$\sf\huge\bold{Answer:}$

Given :

Qadratic polynomial

2x²+x-3

To find :

Zeroes of 2x²+x-3

To verify :

Relation between zeroes and coefficient

Solution :

$\fbox{Split\:Middle\:Term}$

$\tt\ = 2 {x}^{2} + x - 3$

$\tt\ = 2 {x}^{2} - 3x + 2x - 3$

$\tt\ = 2 {x}^{2} + 2x-3x - 3$

$\tt\ = 2x(x + 1) - 3(x + 1)$

$\tt\ = (2x - 3)(x + 1)$

• To find zeroes put polynomial = 0

$\tt\ 2 {x}^{2} + x - 3 = 0$

$\tt\ \: (2x - 3)(x + 1) = 0$

$\tt\ x = \frac{ 3}{2} \: ,</p><p> \: -1$

$\boxed{\bold\purple{x = \frac{3}{2} ,-1}}$

Verification :

There are two relations between zeroes and coefficient

1. Sum of zeroes
2. Product of zeroes

$\boxed{\tt \red {sum \: of \: zeroes( \alpha + \beta ) = \frac{coefficient \: of \: {x}(-b) }{coefficient \: of \: {x}^{2}(a) } }}$

• α = 3/2
• β = - 1
• a = 2
• b = 1

$\tt\ :⟶ \frac{3}{2} +(- 1 )= \frac{1}{2}$

$\tt\ :⟶ \: \frac{ 3 - 4}{2} = \frac{1}{2}$

$\tt\ :⟶ \: \frac{-1}{2} = \frac{-1}{2}$

$\bf\purple{LHS=RHS}$

Hence verified.

$\boxed{\tt \red {product \: of \: zeroes( \alpha \beta ) = \frac{\:constant \: term \: (c) }{coefficient \: of \: {x}^{2}(a) } }}$

• α = 3/2
• β = -1
• c = -3
• a = 2

$\tt\ :⟶ \frac{3}{2} × (-1) = \frac{-3}{2}$

$\tt\ :⟶ \: \frac{ - 3}{2} = \frac{-3}{2}$

$\bf\purple{LHS=RHS}$

Hence, verified .