Question

2
Find the value of k for which quadratic equation has real and equal roots : 4x^2+Kx+9=0
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Answer 1

EXPLANATION.

Quadratic equation.

⇒ 4x² + kx + 9 = 0.

As we know that,

⇒ D = Discriminant Or b² - 4ac.

Roots are real and equal : D = 0.

⇒ (k)² - 4(4)(9) = 0.

⇒ k² - 144 = 0.

⇒ k² = 144.

⇒ k = √144.

⇒ k = 12.

                                                                                                                         

MORE INFORMATION.

Conjugate roots.

(1) = If D < 0.

One roots = α + iβ.

Other roots = α - iβ.

(2) = If D > 0.

One roots = α + √β.

Other roots = α - √β.

Answer 2

Answer:

Given :-

• A quadratic equation has real and equal roots of 4x² + kx + 9 = 0.

To Find :-

• What is the value of k.

Formula Used :-

$\clubsuit$ Discriminate Formula :

$\longmapsto \sf\boxed{\bold{\pink{Discriminate\: (D) =\: b^2 - 4ac}}}$

Solution :-

Given equation :

$\bigstar\: \: \: \sf\bold{\purple{4x^2 + kx + 9 = 0}}$

where,

• a = 4
• b = k
• c = 9

According to the question by using the formula we get,

$\longrightarrow \sf Discriminate\: (D) =\: (k)^2 - 4(4)(9) =\: 0$

$\longrightarrow \sf Discriminate\: (D) =\: (k)^2 - 4 \times 4 \times 9 =\: 0$

$\longrightarrow \sf Discriminate\: (D) =\: (k)^2 - 16 \times 9 =\: 0$

$\longrightarrow \sf Discriminate\: (D) =\: (k)^2 =\: (k)^2 - 144 =\: 0$

$\longrightarrow \sf Discriminate\: (D) =\: (k)^2 =\: 144$

$\longrightarrow \sf Discriminate\: (D) =\: k =\: \sqrt{144}$

$\longrightarrow \sf\bold{\red{Discriminate\: (D) =\: k =\: 12}}$

$\therefore$ The value of k is 12.