Math, asked by enakshipathak24, 1 year ago

Please answer these questions .a,b,c and d are the options . Please show me how to do the sums step by step . I will mark your answer BRAINLIEST.

Attachments:

Answers

Answered by Rohit18Bhadauria
6

6.

To Find:

  • A quadratic equation whose roots are 3 and -3

Solution:

We know that,

A general quadratic is of the form of

\sf{x^{2}-(Sum\:of\:Zeroes)x+Product\:of\:Zeroes=0}

So, Here

Sum of Zeroes= 3+(-3)= 3-3= 0

Product of Zeroes= 3×(-3)= -9

Now,

Required quadratic equation is

✏ x²-(0)x+(-9)= 0

✏ x²-9= 0

Correct answer is option (a)

\rule{190}{2}

7. Correct Question:

Roots of \bf{-x^{2}+\dfrac{1}{2}x+\dfrac{1}{2}=0} is

To Find:

Roots of \bf{-x^{2}+\dfrac{1}{2}x+\dfrac{1}{2}=0}

Solution:

We know that,

Discriminant of the quadratic equation of the form ax²+bx+c=0 is

\pink{\boxed{\sf{D=b^{2}-4ac}}}

Now, given quadratic equation is

\sf{-x^{2}+\dfrac{1}{2}x+\dfrac{1}{2}=0}

\sf{x^{2}-\dfrac{1}{2}x-\dfrac{1}{2}=0}

Now, discriminant of given quadratic equation is

\longrightarrow\sf{D=\bigg(\dfrac{-1}{2}\bigg)^{2}-4(1)\bigg(\dfrac{-1}{2}\bigg)}

\longrightarrow\sf{D=\dfrac{1}{4}+2}

\longrightarrow\sf{D=\dfrac{1+8}{4}}

\longrightarrow\sf{D=\dfrac{9}{4}}

Also, we know that

Roots of Quadratic equation is

\sf{x=\dfrac{-b\pm\sqrt{D}}{2a}}

\sf{x=\dfrac{-\bigg(-\dfrac{1}{2}\bigg)\pm\sqrt{\dfrac{9}{4}}}{2(1)}}

\sf{x=\dfrac{\dfrac{1}{2}\pm\dfrac{3}{2}}{2}}

\sf{x=\dfrac{\dfrac{1}{2}+\dfrac{3}{2}}{2},\dfrac{\dfrac{1}{2}-\dfrac{3}{2}}{2}}

\sf{x=\dfrac{\dfrac{1+3}{2}}{2},\dfrac{\dfrac{1-3}{2}}{2}}

\sf{x=\dfrac{1+3}{4},\dfrac{1-3}{4}}

\sf{x=\dfrac{4}{4},\dfrac{-2}{4}}

\sf{x=1,\dfrac{-1}{2}}

The correct answer is option (a)

\rule{190}{2}

8.

Given:

An equation kx(x-2)+6=0 having equal roots

To Find:

  • Value of k

Solution:

We know that, condition for a quadratic equation of the form  ax²+bx+c=0 to have equal roots is

\bf{b^{2}-4ac=0}

Now, the given equation is

kx(x-2)+6= 0

kx²-2kx+6= 0

It is given that above equation has equal roots

So,

\longrightarrow\sf{(-2k)^{2}-4(k)(6)=0}

\longrightarrow\sf{4k^{2}-24k=0}

\longrightarrow\sf{4(k^{2}-6k)=0}

\longrightarrow\sf{k^{2}-6k=0}

\longrightarrow\sf{k(k-6)=0}

\longrightarrow\sf{k=0,6}

Hence, the value of k are 0 and 6.

\rule{190}{2}

9.

To Find:

  • A quadratic equation whose roots are \sf{2+\sqrt{5}} and \sf{2-\sqrt{5}}

Solution:

We know that,

A general quadratic is of the form of

\sf{x^{2}-(Sum\:of\:Zeroes)x+Product\:of\:Zeroes=0}

So, Here

\sf{Sum\:of\:Zeroes=2+\sqrt{5}+2-\sqrt{5}=4}

\sf{Product\:of\:Zeroes=(2+\sqrt{5})\times(2-\sqrt{5})}

\sf{Product\:of\:Zeroes=4-5=-1}

Now,

Required quadratic equation is

✏ x²-(4)x+(-1)= 0

✏ x² -4x-1= 0

Correct answer is option (a)

\rule{190}{2}

10.

Given:

A quadratic equation 3x²+5x-7=0 whose roots are α and β.

To Find:

  • Value of αβ

Solution:

We know that,

For a quadratic equation

\sf{Product\:of\:Zeroes=\dfrac{Constant\:term}{Coefficient\:of\:x^{2}}}

So, here

\sf{Product\:of\:Zeroes=\alpha\beta}

\sf{Product\:of\:Zeroes=\dfrac{-7}{3}}

From above two equations,

\longrightarrow\sf{\alpha\beta=\dfrac{-7}{3}}

Hence, the correct option is (b).

Answered by asritadevi2emailcom
0

Step-by-step explanation:

 =  >  \alpha  \beta  =  \frac{ - 7}{3}  \\ is \:  \: the \:  \: correct \:  \: answer.

Similar questions