For each of the quadratic equations below, convert them into the standard form $a x^{2} + b x + c = 0$ and determine the value of ' $b^{2} - 4 a c$ . Comment on the nature of roots from the value of ' $b^{2} - 4 a c$
(i) $5 x^{2} = 6 x - 7$
(ii) $4 x^{2} + 16 = - 16 x$
(iii) $2 x^{2} + 6 x + 3 = 0$

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Solution

$i) 5 x^{2} - 6 x + 7 = 0$
$b^{2} - 4 a c = 6^{2} - 4 \cdot 5 \cdot 7$
$= 36 - 140$
$= - 104$ , No real roots
$i i) 4 x^{2} + 16 x + 16 = 0$
$b^{2} - 4 a c = 16^{2} - 4 \cdot 4 \cdot 16$
$= 256 - 256$
$= 0$ , Equal real roots
$i i i) 2 x^{2} + 6 x + 3 = 0$
$b^{2} - 4 a c = 6^{2} - 4 \cdot 2 \cdot 3$
$= 36 - 24$
$= 12$ , Two distinct real roots.

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Similar questions

Q. What is the nature of the roots of the quadratic equation

a x^{2} + b x + c = 0

if
(a)

b^{2} - 4 a c = 0

(b)

b^{2} - 4 a c < 0

Q. If

b^{2} - 4 a c \geq 0

, then write the nature of roots of a quadratic equation

a x^{2} + b x + c = 0

The roots of a quadratic equation $a x^{2} + b x + c = 0$ are given by $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ , provided $b^{2} - 4 a c \geq 0$ .

Q. STATEMENT -1 : Roots of the quadratic equation

3 x^{2} - 2 \sqrt{6} x + 2 = 0

are same.
STATEMENT -2 : A quadratic equation

a x^{2} + b x = c = 0

has two distinct real roots, if

b^{2} - 4 a c > 0

Q. If

b^{2} - 4 a c \geq 0

then the roots of quadratic equation

a x^{2} + b x + c = 0

is-

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For each of the quadratic equations below, convert them into the standard form ax2+bx+c=0 and determine the value of 'b2−4ac. Comment on the nature of roots from the value of 'b2−4ac(i) 5x2=6x−7(ii) 4x2+16=−16x(iii) 2x2+6x+3=0

i)5x2−6x+7=0b2−4ac=62−4⋅5⋅7=36−140=−104, No real rootsii)4x2+16x+16=0b2−4ac=162−4⋅4⋅16=256−256=0, Equal real rootsiii)2x2+6x+3=0b2−4ac=62−4⋅2⋅3=36−24=12, Two distinct real roots.

For each of the quadratic equations below, convert them into the standard form $a x^{2} + b x + c = 0$ and determine the value of ' $b^{2} - 4 a c$ . Comment on the nature of roots from the value of ' $b^{2} - 4 a c$
(i) $5 x^{2} = 6 x - 7$
(ii) $4 x^{2} + 16 = - 16 x$
(iii) $2 x^{2} + 6 x + 3 = 0$