Question

If the roots of the equation X square - 6 X + 10 is equal to zero are alpha and beta then Alpha square plus beta square is

Answer 1

EXPLANATION.

α and β are the roots of the equation.

⇒ x² - 6x + 10 = 0.

As we know that,

Sum of the zeroes of the quadratic polynomial.

⇒ α + β = - b/a.

⇒ α + β = -(-6)/1 = 6.

Products of the zeroes of the quadratic polynomial.

⇒ αβ = c/a.

⇒ αβ = (10)/1 = 10.

To find : α² + β².

⇒ α² + β² = [α + β]² - 2αβ.

Put the values in the equation, we get.

⇒ α² + β² = [6]² - 2(10).

⇒ α² + β² = 36 - 20.

⇒ α² + β² = 16.

                                                                                                                     

MORE INFORMATION.

Nature of the roots of the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

Answer 2

Corrected question:

• If the roots of the equation x² - 6x + 10 is equal to zero are α and β then α² + β² is.

Information provided with us:

• Equation is x² - 6x + 10
• The given equation roots are equal to zero are alpha and beta.

Need to be calculated:

• Alpha square plus beta square

Let's start:

The standard form of a quadratic equation is,

• $\tt{ax {} ^{2} + bx + c = 0 }$

As the given equation is x² - 6x + 10.

Here,

• a is 1
• b is 6
• c is 10

Finding out sum of zeroes:-

‎ ‎ ‎ ‎ ‎ ‎ _______________

Using Formula,

Sum of zeroes of polynomial:-

• $\red{ \underline{{\boxed{\sf{ \alpha + \beta \: = \: \dfrac{ - b}{a} }}}}}$

Here, we have b is 6 and a is 1.

Substituting the values we get,

$: \longmapsto \: \sf{ \alpha + \beta \: = \: \dfrac{ - (-6 )}{1} }$

$: \longmapsto \: \sf{ \alpha + \beta \: = \: \dfrac{ 6}{1} }$

$: \longmapsto \: \boxed{\sf{ \alpha + \beta \: = \: 6 }}$

$\therefore \: \underline{ \sf{Sum \: of \: zeroes \: of \: polynomials \: is \: 6}}$

Finding out product of the zeroes:-

‎ ‎ ‎ ‎ ‎ ‎ _______________

Using Formula,

Product of zeroes of a quadratic polynomial is calculated by,

• $\red{ \underline{\boxed{ \sf{ \alpha \beta \: = \: \dfrac{c}{a} }}}}$

Here we have,

• c is 10
• a is 1

Substituting the values we get,

$:\longmapsto \: \sf{ \alpha \beta \: = \: \dfrac{10}{1} }$

$:\longmapsto \: \boxed{ \sf{ \alpha \beta \: = \: 10}}$

$\therefore \: \underline{ \sf{Product\: of \: zeroes \: of \: polynomials \: is \: 10}}$

Finding out α² + β²:-

Using Identity,

• $\red{ \underline{\boxed{ \sf{ a {}^{2} + b {}^{2} = \: ( a + b) {}^{2} - 2 ab }}}}$

Here we have,

• a + b is 6 (i.e, α + β)
• a is 2 (i.e, α)
• b is 10 (i.e., β)

Substituting the values we get,

$\longrightarrow \: \sf{ \alpha {}^{2} + \beta {}^{2} = \: ( 6 ) {}^{2} - 2 (10) }$

$\longrightarrow \: \sf{ \alpha {}^{2} + \beta {}^{2} = \: ( 6 \times 6)- 2 (10) }$

$\longrightarrow \: \sf{ \alpha {}^{2} + \beta {}^{2} = \: 36 \: - \: 2 (10) }$

$\longrightarrow \: \sf{ \alpha {}^{2} + \beta {}^{2} = \: 36 \: - \: 2 \times 10 }$

$\longrightarrow \: \sf{ \alpha {}^{2} + \beta {}^{2} = \: 36 \: - \: 20 }$

$\longrightarrow \: \underline{\boxed{\sf{ \alpha {}^{2} + \beta {}^{2} = \: 16}}}$

• $\underline{\bf{Hence, \: \alpha {}^{2} \: + \: \beta {}^{2} \: is \: 16 \: }}$