Question

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.

Answer 1

All of your five questions are solved in 3 pages...have a look...!!!

Answer 2

1.

To Find:

Quadratic equation among given equation.

Solution:

We know that,

Those polynomials can be called quadratic in which highest degree or power of x is 2. x² must be present in the equation.

Now, in the given equations

On simplifying all the equations, we came to know that only equation (b) is the quadratic equation because highest power of x is 2 here

While in (a)- Highest power of x is 3

While in (c)- Highest power of x is 4

While in (d)- Highest power of x is 3

Hence, (b) is the correct answer

2. Correct Question:

Value of x for the equation $\sf{a^{2}x^{2}-3abx+2b^{2}=0}$ is:

Given:

A quadratic equation $\bf{a^{2}x^{2}-3abx+2b^{2}=0}$

To Find:

Value of x

Solution:

$\longrightarrow\sf{a^{2}x^{2}-3abx+2b^{2}=0}$

$\longrightarrow\sf{a^{2}x^{2}-2abx-abx+2b^{2}=0}$

$\longrightarrow\sf{ax(ax-2b)-b(ax-2b)=0}$

$\longrightarrow\sf{(ax-b)(ax-2b)=0}$

$\longrightarrow\sf{x=\dfrac{b}{a},\dfrac{2b}{a}}$

Hence, the correct answer is (a)

3. Correct Question:

Which of the following have rational roots

a) 2x²+x-1=0

b) x²+x+1=0

c) x²-6x+6=0

d) 2x²+15x+30=0

To Find:

Quadratic equation having rational roots among given equations

Solution:

We know that,

• For a quadratic equation a²+bx+c=0 to have rational roots, its discriminant should be equal to 0 or should be the square of number.

• Discriminant= b² -4ac

Now, in the equation (a)

Discriminant= (1)² -4(2)(-1)= 9

And 9 is a square of 3

Hence, (a) is the correct answer

4.

Given:

An equation

$x=\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}$

To Find:

• Value of x

Solution:

$x=\dfrac{1}{2-\dfrac{1}{\dfrac{4-2x-1}{2-x}}}$

$x=\dfrac{1}{2-\dfrac{2-x}{3-2x}}$

$x=\dfrac{1}{\dfrac{6-4x-2+x}{3-2x}}$

$x=\dfrac{3-2x}{4-3x}$

$x(4-3x)=3-2x$

$4x-3x^{2}=3-2x$

$-3x^{2}+4x+2x-3=0$

$-3x^{2}+6x-3=0$

$-3(x^{2}-2x+1)=0$

$x^{2}-2x+1=0$

$(x-1)^{2}=0$

$x-1=0$

$x=1$

Hence, the value of x is 1.

5.

Given:

A quadratic equation $\bf{\sqrt{3}x^{2} +10x+7\sqrt{3}=0}$

To Find:

Solution of given equation by factorisation

Solution:

$\longrightarrow\sf{\sqrt{3}x^{2} +10x+7\sqrt{3}=0}$

$\longrightarrow\sf{\sqrt{3}x^{2} +7x+3x+7\sqrt{3}=0}$

$\longrightarrow\sf{x(\sqrt{3}x +7)+\sqrt{3}(\sqrt{3}x+7)=0}$

$\longrightarrow\sf{(x+\sqrt{3})(\sqrt{3}x +7)=0}$

$\longrightarrow\sf{x=-\sqrt{3}\:,\dfrac{-7}{\sqrt{3}}}$

Hence, the correct answer is option (a).