Question

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2x² + 4x - 1 = 0 has roots α and β form the eqⁿ with α² and β²


Remember :

$: : \leadsto \sf {x}^{2} - ( { \alpha }^{2} + { \beta }^{2})x + { \alpha}^{2} { \beta }^{2} = 0$
​

Answer 1

Answer:-

Given:-

α and β are the roots of 2x² + 4x - 1.

On comparing the given equation with the standard form of a quadratic equation i.e., ax² + bx + c = 0 ;

Let;

• a = 2
• b = 4
• c = - 1.

We know that;

Sum of the roots = - b/a

So,

⟹ α + β = - 4/2

Squaring both sides we get;

⟹ (α + β)² = ( - 2)²

using (a + b)² = a² + b² + 2ab in LHS we get,

⟹ α² + β² + 2αβ = 4

⟹ α² + β² = 4 - 2αβ -- equation (1)

Also;

Product of the roots = c/a

So,

⟹ αβ = - 1/2 -- equation (2)

Substitute αβ = - 1/2 in equation (1).

⟹ α² + β² = 4 - 2( - 1/2)

⟹ α² + β² = 4 + 1

⟹ α² + β² = 5

Squaring both sides in equation (2) we get;

⟹ (αβ)² = ( - 1/2)²

⟹ α²β² = 1/4

Now;

We have:

• Sum of the roots = α² + β² = 5

• Product of the roots = α²β² = 1/4

We know,

Standard form of a quadratic equation is:

⟹ x² - (sum of roots)x + Product of the roots = 0

Hence,

⟹ x² - (5)x + 1/4 = 0

Taking LCM in LHS we get,

⟹ (4x² - 20x + 1)/4 = 0

⟹ 4x² - 20x + 1 = 0

∴ The required quadratic equation is 4x² - 20x + 1 = 0.

Answer 2

Required equation is : 4x^2 - 20 x + 1 = 0

For Explanation refer to attachment :