Question

find the value of K quadratic equation

X²+2√2 Kx+18=0 has equal roots
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Answer 1

★Given:-

• The quadratic equation x²+2√2 Kx+18=0 has equal roots.

★To find:-

• The value of K.

★Solution:-

We know:

If ax²+bx+c = 0 is a quadratic equation,then for it to have equal roots,the discriminant = 0.

$\leadsto \underline{\boxed{\sf b^2 - 4ac=0 }}$

Here,

→a = 1

→b = 2√2k

→c = 18

Putting the values,

$:\implies \sf (2\sqrt{2}k )^2-4(1)(18)=0\\\\:\implies \sf 4\times 2\times k^2 - 72=0\\\\:\implies \sf 8k^2= 72\\\\:\implies \sf k^2=\dfrac{72}{8} \\\\:\implies \sf k^2=9\\\\:\implies \sf k = \sqrt{9} \\\\:\implies \underline{\boxed{\bold{k=3\ or\ -3.}}}$

Hence,

The value of k is  ±3.

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Know More:-

Quadratic formula:

• - b ± √(b² − 4ac) / 2a

→ b² − 4ac is also called as determinant.

If,

→ b² − 4ac = 0 - Real & equal roots

→ b² − 4ac > 0 - Real & unequal roots

→ b² − 4ac < 0 - Imaginary numbers

________________

Answer 2

★Given:-

The quadratic equation x²+2√2 Kx+18=0 has equal roots.

★To find:-

The value of K.

★Solution:-

We know:

If ax²+bx+c = 0 is a quadratic equation,then for it to have equal roots,the discriminant = 0.

$\leadsto \underline{\boxed{\sf b^2 - 4ac=0 }}$

Here,

→a = 1

→b = 2√2k

→c = 18

Putting the values,

$:\implies \sf (2\sqrt{2}k )^2-4(1)(18)=0\\\\:\implies \sf 4\times 2\times k^2 - 72=0\\\\:\implies \sf 8k^2= 72\\\\:\implies \sf k^2=\dfrac{72}{8} \\\\:\implies \sf k^2=9\\\\:\implies \sf k = \sqrt{9} \\\\:\implies \underline{\boxed{\bold{k=3\ or\ -3.}}}$

Hence,

The value of k is  ±3.

-----------------------------

Know More:-

Quadratic formula:

- b ± √(b² − 4ac) / 2a

→ b² − 4ac is also called as determinant.

If,

→ b² − 4ac = 0 - Real & equal roots

→ b² − 4ac > 0 - Real & unequal roots

→ b² − 4ac < 0 - Imaginary numbers

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