Question

Determine whether the given quadratic equation has roots). If so, find the root(s) -2y2 + y + 1 = 0.​

Answer 1

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Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given given a Quadratic Polynomial
• ( - 2y² + y + 1 ) =0

To Find:

• We have to check whether the given quadratic equation has roots or not
• If real roots exists then we have to find the roots

Solution:

We have been given a polynomial

=> - 2y² + y + 1 = 0

=> - (2y² - y - 1) = 0

=> 2y² - y - 1 = 0

Comparing with standard equation

Here a = 2, b = (-1), c = (-1)

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Discriminant = b² - 4ac

Discriminant = (-1)² - 4(2)(-1)

Discriminant (D) = 9

Since it is clear that D > 0

So real and distinct roots exist for given quadratic polynomial.

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Roots of given quadratic equation can be estimated by using Middle Term Splitting

$=> 2{y}^{2} - y - 1 = 0$

$=> 2{y}^{2} - (2-1)y - 1 = 0$

$=> 2{y}^{2} - 2y + y - 1 = 0$

$=> 2y(y-1) + (y-1) = 0$

$=> ( y-1 )(2y + 1 ) = 0$

So Either

$∴ ( y - 1 ) = 0$

$∴ \: y = 1$

OR

$∴ ( 2y + 1 ) = 0$

$∴ \: y = \dfrac{-1}{2}$

Hence the zeros of Quadratic Equation are 1 or - 1/2