Question

if -5 is a root of the quadratic equation 2x^2+px-15=0 and the quadratic equation p(x^2+x)+k=0 has equal roots . find the value of k

Answer 1

Answer:

k = 1.75

Step-by-step explanation:

At first, solving the first part of question to get the value of p.

If (-5) is the root of the equation, it means the remainder should be zero after substituting the value of x = (-5).

⇒ 2x² + px - 15 = 0

⇒ 2(-5)² + p(-5) = 15

⇒ 50 - 5p = 15

⇒ 5p = 50 - 15

⇒ 5p = 35

⇒ p = $\sf \dfrac{35}{5}$

⇒ p = 7

Hence, the value of p is 7.

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Now, on substituting the value of p in the second equation which has equal roots.

⇒ p(x² + x) + k = 0

⇒ 7(x² + x) + k = 0

⇒ 7x² + 7x + k = 0

Here,

• a = 7
• b = 7
• c = k

Now, if it have equal roots, it means Discriminant must be equal to zero.

⇒ Discriminant = 0

⇒ b² - 4ac = 0

⇒ (7)² - 4 * 7 * k = 0

⇒ 49 - 28k = 0

⇒ 28k = 49

⇒ k = $\sf \dfrac{49}{28}$

⇒ k = 1.75

Hence, the value of k is 1.75

Answer 2

$\huge\red{Answer:-}$

(-5) is the root of equation 2x^2+px-15=0

Let us find the value of k

$= 2x {}^{2} + px - 15 = 0$

$= 2( - 5) {}^{2} + p( - 5) = 15$

$= 50 - 5p - = 15$

$= 5p = 50 - 15$

$= 5p = 35$

$= p = \frac{35}{5}$

Let us add the value:

$= p(x {}^{2} + x) + k = 0$

$= 7(x {}^{2} + x) + k = 0$

$= 7x {}^{2} + 7x + k = 0$

After adding the values:

$a = 7$

$b = 7 \: and$

$c = k$

Then,

$b {}^{2} - 4ac = 0$

$(7) {}^{2} - 4..7..$

$k = 0$

$= 49 - 28k = 0$

$= 28 = 49$

$= k = \frac{49}{28}$

$Therefore, \: k = 1.75$