Question

pls answer my question ​

Answer 1

Things to Know before Solving the questions:

For a quadratic polynomial,

• $\bf{Sum\:of\:Zeroes=\dfrac{-Coefficient\:of\:x}{Coefficient\:of\:x^{2}}}$

• $\bf{Product\:of\:Zeroes=\dfrac{Constant\:Term}{Coefficient\:of\:x^{2}}}$

$\rule{180}{2}$

(2)

Given:

• Polynomial p(x)=2x²+px+4 whose one zero is 2

To Find:

• Other zero of polynomial
• Value of p

Solution:

Let the other zero be 'a', then

$\longrightarrow\sf{(2)(a)=\dfrac{4}{2}}$

$\longrightarrow\sf{a=\dfrac{4}{4}}$

$\longrightarrow\sf{a=1}$

Now,

$\longrightarrow\sf{Sum\:of\:Zeroes=\dfrac{-p}{2}}$

$\longrightarrow\sf{2+1=\dfrac{-p}{2}}$

$\longrightarrow\sf{3=\dfrac{-p}{2}}$

$\longrightarrow\sf{p=-3\times2}$

$\longrightarrow\sf\pink{p=-6}$

Hence, other zero is 1 and value of p is -6.

$\rule{180}{2}$

(3)

Given:

• Polynomial x²-p(x+1)+c whose zeroes are α and β
• (α+1)(β+1)=0

To Find:

• Value of c

Solution:

Given polynomial,

➺ x²-p(x+1)+c

➺ x²-px-p+c

Now,

$\longrightarrow\sf{Sum\:of\:Zeroes=\dfrac{-(-p)}{1}}$

$\longrightarrow\sf{\alpha+\beta=p------(1)}$

Also,

$\longrightarrow\sf{Product\:of\:Zeroes=\dfrac{-p+c}{1}}$

$\longrightarrow\sf{\alpha\beta=-p+c------(2)}$

Now, it is given that

➳ (α+1)(β+1)=0

➳ α(β+1)+1(β+1)=0

➳ αβ+α+β+1=0

From (1) and (2)

➳ -p+c+p+1=0

➳ c+1=0

➳ c= -1

Hence, value of c is -1.

$\rule{180}{2}$

(4)

Given:

• Polynomial 3x²-2(4x+k)+1
• One zero of given polynomial is 7 times of other

To Find:

• Both the zeroes of the polynomial
• Value of 'k'

Solution:

Given polynomial,

➺ 3x² -2(4x+k)+1

➺ 3x² -8x-2k+1

Let one zero be α and other zero be β, where α > β

So,

α= 7β

Now,

$\longrightarrow\sf{Sum\:of\:Zeroes=\dfrac{-(-8)}{3}}$

$\longrightarrow\sf{\alpha+\beta=\dfrac{8}{3}------(1)}$

$\longrightarrow\sf{7\beta+\beta=\dfrac{8}{3}}$

$\longrightarrow\sf{8\beta=\dfrac{8}{3}}$

$\longrightarrow\sf{\beta=\dfrac{1}{3}}$

On putting value of β in (1), we get

$\longrightarrow\sf{\alpha+\dfrac{1}{3}=\dfrac{8}{3}}$

$\longrightarrow\sf{\alpha=\dfrac{8}{3}-\dfrac{1}{3}}$

$\longrightarrow\sf{\alpha=\dfrac{8-1}{3}}$

$\longrightarrow\sf{\alpha=\dfrac{7}{3}}$

Also,

$\longrightarrow\sf{Product\:of\:Zeroes=\dfrac{-2k+1}{3}}$

$\longrightarrow\sf{\alpha\beta=\dfrac{-2k+1}{3}}$

$\longrightarrow\sf{\dfrac{7}{3}\times\dfrac{1}{3} =\dfrac{-2k+1}{3}}$

$\longrightarrow\sf{\dfrac{7\times3}{3\times3}=-2k+1}$

$\longrightarrow\sf{\dfrac{7}{3}=-2k+1}$

$\longrightarrow\sf{7=3(-2k+1)}$

$\longrightarrow\sf{7=-6k+3}$

$\longrightarrow\sf{6k=3-7}$

$\longrightarrow\sf{6k=-4}$

$\longrightarrow\sf{k=\dfrac{-4}{6}}$

$\longrightarrow\sf{k=\dfrac{-2}{3}}$

Hence, both zeroes are  $\sf{\dfrac{7}{3}}$ and $\sf{\dfrac{1}{3}}$ and value of k is $\sf{\dfrac{-2}{3}}$.

$\rule{180}{2}$

(5)

Given:

• Polynomial 2x²-5x-(2k+1)
• One zero of given polynomial is twice of other

To Find:

• Both the zeroes of the polynomial
• Value of 'k'

Solution:

Given polynomial,

➺ 2x² -5x-(2k+1)

➺ 2x² -5x-2k-1

Let one zero be α and other zero be β, where α > β

So,

α= 2β

Now,

$\longrightarrow\sf{Sum\:of\:Zeroes=\dfrac{-(-5)}{2}}$

$\longrightarrow\sf{\alpha+\beta=\dfrac{5}{2}------(1)}$

$\longrightarrow\sf{2\beta+\beta=\dfrac{5}{2}}$

$\longrightarrow\sf{3\beta=\dfrac{5}{2}}$

$\longrightarrow\sf{\beta=\dfrac{5}{6}}$

On putting value of β in (1), we get

$\longrightarrow\sf{\alpha+\dfrac{5}{6}=\dfrac{5}{2}}$

$\longrightarrow\sf{\alpha=\dfrac{5}{2}-\dfrac{5}{6}}$

$\longrightarrow\sf{\alpha=\dfrac{15-5}{6}}$

$\longrightarrow\sf{\alpha=\dfrac{10}{6}}$

Also,

$\longrightarrow\sf{Product\:of\:Zeroes=\dfrac{-2k-1}{2}}$

$\longrightarrow\sf{\alpha\beta=\dfrac{-2k-1}{2}}$

$\longrightarrow\sf{\dfrac{10}{6}\times\dfrac{5}{6} =\dfrac{-2k-1}{2}}$

$\longrightarrow\sf{\dfrac{50}{36}=\dfrac{-2k-1}{2}}$

$\longrightarrow\sf{\dfrac{50\times2}{36}=-2k-1}$

$\longrightarrow\sf{\dfrac{50}{18}=-2k-1}$

$\longrightarrow\sf{50=18(-2k-1)}$

$\longrightarrow\sf{50=-36k-18}$

$\longrightarrow\sf{36k=-50-18}$

$\longrightarrow\sf{36k=-68}$

$\longrightarrow\sf{k=\dfrac{-68}{36}}$

$\longrightarrow\sf{k=\dfrac{-17}{9}}$

Hence, both zeroes are  $\sf{\dfrac{10}{6}}$ and $\sf{\dfrac{5}{6}}$ and value of k is $\sf{\dfrac{-17}{9}}$.