If $b^{2} - 4 a c > 0$ , then the graph of $y = a x^{2} + b x + c$

cuts x-axis in two real points.

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touches the x-axis.

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lies entirely above the x-axis.

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can not be determined.

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Solution

The correct option is A cuts x-axis in two real points.
$a x^{2} + b x + c = 0$

The roots of the above equation are given by the quadratic formula
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
It is given that
$b^{2} - 4 a c > 0$
Hence
$D > 0$ .
Therefore, we will get two distinct real roots.
This implies that, the graph of y vs x cuts the x axis at two distinct real points.

Suggest Corrections

Similar questions

a < 0

D < 0

y = a x^{2} + b x + c

(where D = b^{2} - 4 a c)

b^{2} - 4 a c = 0

y = a x^{2} + b x + c

a x^{2} + b x + c

c = \frac{b^{2}}{4 a}

a > 0

b^{2} - 4 a c > 0

y = a x^{2} + b x + c

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