Question

Find the zeros of the quadratic polynomial f(x) = 6x²-3, and verify the relationship between the zeros and its coefficients.​

Answer 1

Explanation:

1) zeros of the given function:

6x²-3=0; ⇔ 6(x²-0.5)=0; ⇔ x²=0.5; ⇔

`\left[\begin{array}{ccc}x=-√(0.5) \\x=√(0.5) \end{array}`

2) relationship:

if to see the equation x²-0.5=0 (ax²+bx+c=0 is standart form!), then the sum of the zeros is '0' (it is 'b' of the standart form), the product of equation roots is '-0.5' (it is 'c' of the standart form).

Answer 2

`{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}`

★ The polynomial

f(x) = 6x² - 3

`{\large{\textsf{\textbf{\underline{\underline{To \: find :}}}}}}`

★ Zeroes of the polynomial f(x) = 6x² - 3

`{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}`

We have,

`f(x) = \tt 6 {x}^(2) - 3`

Which can also be written as

`\implies f(x) = \tt {(√(6) x)}^(2) - { (√(3)) }^(2)`

Using a² - b² = (a - b) (a + b)

`\implies f(x) = \tt ( √(6) x - √(3) )( √(6) x + √(3) )`

To find the zeroes, solve f(x) = 0

`\longrightarrow \tt ( √(6) x - √(3) )( √(6) x + √(3) ) = 0`

either
`\tt √(6) x - √(3) = 0 \: or \: √(6) x + √(3) = 0`

`\implies \tt √(6) x = √(3 \: ) \: or \: \: √(6) x = - √(3)`

`\implies \tt x = ( √(3) )/( √(6) ) \: or \: x = - ( √(3) )/( √(6) )`

`\implies \tt x = ( √(3) )/( √(2 * 3) ) \: or \: x = - ( √(3) )/( √(2 * 3) )`

`\implies \tt x = \frac{ \cancel{ √(3) }}{ √(2) \: \cancel{√(3)} } \: or \: x = - \frac{ \cancel{ √(3) }}{ √(2) \: \cancel{√(3)} }`

`\implies \tt x = (1)/( √(2) ) \: \: or \: \: - (1)/( √(2) )`

Hence, the zeroes of f(x) = 6x² - 3 are:

`\tt \alpha =\sf \boxed {{ \red{ (1)/( √(2) ) } }}\: \: and \: \: \beta =\sf \boxed {{ \red{ - (1)/( √(2) ) } }}`

• Verification

Sum of zeroes =
`( \alpha + \beta )`

`= \tt (1)/( √(2) ) + \bigg(- (1)/( √(2) ) \bigg)`

`= \tt (1)/( √(2) ) + - (1)/( √(2) )`

`= \tt 0`

and,
`\tt - \frac{Coefficient \: of \: x}{Coefficient \: of \: {x}^(2) }`

`\tt = - (0)/(6)`

`\tt = 0`

`\therefore \tt \: Sum \: of \: zeroes = {\boxed{ \red{\frac{ \tt Coefficient \: of \: x}{ \tt Coefficient \: of \: {x}^(2)}}}}`

Also,

Product of zeroes =
`\alpha \beta`

`= (1)/( √(2) ) * - (1)/( √(2) )`

`= - (1)/( 2 )`

and,
`\tt - \frac{Constant \: term}{Coefficient \: of \: {x}^(2) }`

`\tt = ( - 3)/(6)`

`\tt = ( - 1)/(2)`

`\therefore \tt \: Product \: of \: zeroes = {\boxed{ \red{\frac{ \tt Constant \: term}{ \tt Coefficient \: of \: {x}^(2)}}}}`

`\rule{280pt}{2pt}`