Question

1) if -4 is a root of the equation x^2 + Px - 4 = 0 and the equation x^2 + Px + q =0 has equal roots , find the value of p and q
2) if -5 is a root of the equation 2x^2 +Px -15 = 0 and the equation p(x^2 + x) + K = 0
has equal roots, finds the value of k.

Answer 1

Answer:

here is your answer and once try to solve it yourself

Answer 2

AnswEr 1 :

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

If -4 is a root of the equation x² + px - 4 = 0 and the equation x² + px + q =0 has equal roots.

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

The value of p and q.

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

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

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

Now;

x² + px + q=0 has equal roots.

• a = 1
• b = p(3)
• c = q

$\mapsto\sf{Discriminant\:(D)=b^{2} -4ac=0}\\\\\mapsto\sf{(3)^{2} -4\times 1\times q=0}\\\\\mapsto\sf{9-4q=0}\\\\\mapsto\sf{9=4q}\\\\\mapsto\sf{\red{q=\dfrac{9}{4}}}$

∴ The value of p and q is 3 and 9/4.

_______________________________________________

AnswEr 2 :

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

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.

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

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

Now;

p(x² + x) + k = 0

px² + px + k = 0

7x² + 7x + k = 0

• a = 7
• b = 7
• c = k

A/q

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

Thus,

The value of k is 7/4 .