Question

answer this question​

Answer 1

21(x) x 61(w)

the answer is that. it is simple once you get the hang of it

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Answer 2

$\huge\boxed{\textsf{\textbf{\color{magenta}{QuesTion:-}}}}$

For which value of k, the quadratic equation kx²-3x+k=0 has real and equal roots ?

$\huge\boxed{\textsf{\textbf{\color{cyan}{AnsWer:-}}}}$

• Equation : kx² - 3x + k = 0
• a = k
• b = -3
• c = k

Discriminant = b² - 4ac

Since, it has equal and real roots, b² - 4ac = 0

==> ( -3 )² - 4 ( k ) ( k ) = 0

==> 9 - 4k² = 0

==> - 4k² = -9

==> - k² = $\frac{-9}{4}$

==> - k = $\frac{√-9}{√4}$

==> k = $\frac{-3}{2}$

Therefore, for k = -³/2 the equation kx² - 3x + k = 0 has equal and real roots.

☔VeriFication:-

$\frac{-3}{2}x²-3x+\frac{-3}{2}=0$

==> -3x² - 6x - 3 = 0

==> Discriminant = b²-4ac ( For equal and real roots )

==> ( -6 )² - 4 ( -3 ) ( -3 ) = 0

==> 36 - 36 = 0

==> 0 = 0

Hence verified !!

$\huge\boxed{\textsf{\textbf{\color{magenta}{QuesTion:-}}}}$

If LCM (X,Y) = 78 and the product of X and Y is 1560. Find its HCF.

$\huge\boxed{\textsf{\textbf{\color{cyan}{AnsWer:-}}}}$

$LCM×HCF=X×Y$

• LCM of (X and Y) = 78
• Product of X and Y = 1560
• HCF = ?

==> 78 × HCF = 1560

==> HCF = $\frac{\cancel1\cancel5\cancel6\cancel0}{\cancel7\cancel8}$

==> HCF = 20

Therefore, the HCF is 20 if LCM (X,Y) = 78 and product of X and Y is 1560.

☔VeriFication:-

LCM (X,Y) × HCF (X,Y) = Product of X and Y

==> 78 × 20 = 1560

==> 1560 = 1560

Hence verified !!

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Hope it helps you

Nani...♥️✌️

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