Question

the discriminal of quadratic equation x²-2x+1= 0 of the value will be:

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Answer 1

The quadratic formula states:

For

$\bf \huge \pink{ {ax}^{2} + bx + c = 0}$

the values of x

which are the solutions to the equation are given by:

$\bf \huge \fbox \green{ \: \: x = \frac{ - b ±\sqrt{ {b}^{2} - 4ac } }{2a} \: \: }$

The discriminate is the portion of the quadratic equation within the radical:

$\bf \huge \orange{ {b}^{2} - 4ac}$

If the discriminate is:

- Positive, you will get two real solutions

- Zero you get just ONE solution

- Negative you get complex solutions

To find the discriminant for this problem substitute:

a=1

b=2

c=-1

$\bf \green{ \implies {(2)}^{2} - 4(1)( - 1)}$

$\bf \green{ \implies4 + 4}$

$\bf \green{ \implies8}$

Therefore, this quadratic would have two real solutions.

Answer 2

$\sf{\underline{\boxed{\green{\large{\bold{ Question}}}}}}$

• solve the equation using formulae $\sf x^2 - 2x + 1 = 0$

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$\sf{\underline{\boxed{\green{\large{\bold{ Solution}}}}}}$

$\sf\implies x^2 - 2x + 1 = 0$

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• compare the eq with $\sf{\underline{\bold{ax^2 + bx + c = 0 }}}$

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☯ a = 1

☯ b = -2

☯ c = +1

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• now :-

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$\sf{\underline{\boxed{\pink{\large{\mathfrak{ D = b^2 - 4ac }}}}}}$

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• finding value of D.

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$\sf\implies D = b^2 - 4ac$

$\sf\implies D = (-2)^2 - 4 \times 1 \times -1$

$\sf\implies D = 4 - 4$

$\sf\implies D = 0$

$\sf{\underline{\boxed{\blue{\large{\bold{ D = 8}}}}}}$