Question

The value of k, so that 16x ^ 2 + kx + 1 = 0 has equal roots
1) 48
2)-8
3) 18
4) 14​

Answer 1

Answer:

2) -8

Explanation:

Given equation,

$\rm \implies \: {16x}^{2} + kx + 1 = 0$

On comparing with the general form of a quadratic equation ax² + bx + c = 0

We get,

• a = 16
• b = k
• c = 1.

Finding ‘D’ :-

$\rm \implies \: D = \sqrt{{b}^{2} - 4ac}$

$\rm \implies \: D= \sqrt{{k}^{2} - 4(16)(1)}$

$\rm \implies \: D= \sqrt{{k}^{2} - 64}$

We know that,

For equal roots the discriminant D must be equal to zero.

$\rm \implies \: \sqrt{{k}^{2} - 64} = 0$

Squaring both sides,

$\rm \implies k^2 - 64 = 0$

$\rm \implies k^2 = 64$

$\rm \implies k = \pm\sqrt{64}$

$\rm \implies k = 8, -8$

Required answer: -8

____________________________

Note:

• D(discriminant) = $\rm \sqrt{b^2 - 4ac}$
• When the roots of an equation are equal D = 0.

Answer 2

Answer:

Given :-

• 16x² + kx + 1 = 0 has equal roots.

To Find :-

• What is the value of k.

Formula Used :-

$\bigstar$ Discriminant Formula :

$\dashrightarrow \sf\boxed{\bold{\pink{Discriminant =\: b^2 - 4ac}}}$

Solution :-

Given Equation :

$\mapsto \bf 16x^2 + kx + 1 =\: 0$

By comparing with ax² + bx + c = 0 we get,

⦿ a = 16

⦿ b = k

⦿ c = 1

According to the question by using the formula we get,

$\implies \sf\bold{\green{Discriminant =\: b^2 - 4ac}}$

$\implies \sf Discriminant =\: (k)^2 - 4(16)(1)$

$\implies \sf Discriminant =\: k^2 - 4 \times 16 \times 1$

$\implies \sf Discriminant =\: k^2 - 4 \times 16$

$\implies \sf\bold{\purple{Discriminant =\: k^2 - 64}}$

Now, the given equation will have real and equal roots :

$\implies \bf Discriminant =\: 0$

$\implies \sf k^2 - 64 =\: 0$

$\implies \sf k^2 =\: 64$

$\implies \sf k =\: \sqrt{64}$

$\implies \sf k =\: \sqrt{\underline{8 \times 8}}$

$\implies \sf k =\: 8$

$\implies \sf\bold{\red{k =\: ±\: 8}}$

$\therefore$ The value of k is ± 8 .

Hence, the value of k, according to the option is - 8 .

So, the correct options is (2) - 8 .