Question

Q1.For ax2 + bx + c = 0, which of the following statement is wrong? (a) If b2 – 4ac is a perfect square, the roots are rational. (b) If b2 = 4ac , the roots are real and equal. (c) If b2 – 4ac is negative, no real roots exist. (d) If b2 = 4ac , the roots are real and unequal.

Answer 1

Step-by-step explanation:

For ax² + bx + c = 0,

The roots are given by

$x = \frac{ - b \: \pm \: \sqrt{ { b}^{2} - 4ac } }{2a}$

If b² - 4ac is perfect square, then

$\sqrt{b ^{2} - 4ac}$

is rational.

Therefore, We will have rational roots.

If b² = 4ac then,

b² - 4ac = 0

$\begin{gathered}x = \frac{ - b \: \pm \: \sqrt{ { b}^{2} - 4ac } }{2a} \\ \\ x = \frac{ - b \pm \: 0}{2a} \\ \\ x = \frac{ - b}{2a} \end{gathered}$

Therefore, The roots are real and equal.

If b² - 4ac < 0, then

$\sqrt{ {b}^{2} - 4ac }$

is irrational.

So no real roots exist in this case.

Therefore, A, B, C Options are correct. Wrong statement is Option D, If b² = 4ac , the roots are real and unequal.

Answer 2

Answer:

Answer: D)

as we know the nature of roots is determined by it's discriminant

i.e D = b²-4ac

as b²= 4ac : D = 0

so when we take the sqare root of zero we get it as zero only

so the zeros of the polynomial will be -b/2a

so we can conclude that is the d= 0 the zeros of the polynomial are real and equal

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