Question

5. 'p(x^(2) x) q=0' is a root of the quadratic equation 'dots..' . For '2x^(2) px-15=0' value ofq, the equation '-3' has equal roots.​

Answer 1

$\huge\orange{\underline\blue{\underline\green{\underline{ \mathcal\orange{Ans}\mathcal\green{wer}}}}}❤️$

-5 is a root of quadratic equation 2x² + px - 15 = 0

so, 2(-5)² + p(-5) - 15 = 0

=> 50 - 5p - 15 = 0

=> 35 - 5p = 0

p = 7

now, put p = 7 in second quadratic equation,

p(x² + x ) + k = 0

=> 7(x² + x ) + k = 0

=> 7x² + 7x + k = 0

above equation has equal roots

so, D = b² - 4ac = 0

=> 7² - 4 × 7 × k = 0

=> 7 - 4k = 0

=> k = 7/4 = 1.75

hence, the value of k = 1.75

Answer 2

AnswEr–:

We know that equal roots :

D = 0

★$\bf{\boxed{\sf{b^{2}-4ac}}}$★

We can compared with given above equation ax² + bx + c = 0

∴ x = 4

$\begin{gathered}\mapsto\sf{f(x)=x^{2} +px-4=0}\\\\\mapsto\sf{f(-4)=(-4)^{2} +p(-4)-4=0}\\\\\mapsto\sf{f(-4)=16+(-4p)-4=0}\\\\\mapsto\sf{f(-4)=16-4p-4=0}\\\\\mapsto\sf{f(-4)=12-4p=0}\\\\\mapsto\sf{f(-4)=12=4p}\\\\\mapsto\sf{f(-4)=p=\cancel{\dfrac{12}{4} }}\\\\\mapsto\sf{\red{p=3}}\end{gathered}$

Now;

x² + px + q=0 has equal roots.

• a = 1
• b = p(3)
• c = 0

________________________________________________

AnswEr :

$\bf{\large{\underline{\underline{\bf{Given\::}}}}}$

If -5 is a root of the equation 2x² + px - 15 = 0 and the equation p(x² + x) + k =0 has equal roots.

$\bf{\large{\underline{\underline{\bf{To\:find\::}}}}}$

• The value of k.

$\bf{\large{\underline{\underline{\bf{Explanation\::}}}}}$

$\begin{gathered}\mapsto\sf{f(x)=2x^{2} +px-15=0}\\\\\mapsto\sf{f(-5)=2(-5)^{2} +p(-5)-15=0}\\\\\mapsto\sf{f(-5)=2*25+(-5p)-15=0}\\\\\mapsto\sf{f(-5)=50-5p-15=0}\\\\\mapsto\sf{f(-5)=35-5p=0}\\\\\mapsto\sf{f(-5)=35=5p}\\\\\mapsto\sf{f(-5)=p=\cancel{\dfrac{35}{5} }}\\\\\mapsto\sf{\red{p=7}}\end{gathered}$

Now;

p(x² + x) + k = 0

px² + px + k = 0

7x² + 7x + k = 0

• a = 7
• b = 7
• c = k
• A/q

$\begin{gathered}\mapsto\sf{D=b^{2} -4ac=0}\\\\\mapsto\sf{(7)^{2} -4\times 7\times k=0}\\\\\mapsto\sf{49-28k=0}\\\\\mapsto\sf{49=28k}\\\\\mapsto\sf{k=\cancel{\dfrac{49}{28} }}\\\\\mapsto\sf{\red{k=\dfrac{7}{5} }}\end{gathered}$

• Thus,

The value of k = 1.75