Answer

Verified

337.8k+ views

**Hint:**We first need to rearrange the given equation in the standard form of quadratic equation. That is in the form of \[a{x^2} + bx + c = 0\]. After that we can solve this using various methods that are by completing the square, factorization, graph or by quadratic formula. Here we need to use a quadratic formula that is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

**Complete step-by-step solution:**

Given, \[7{x^2} - 5 = 2x + 9{x^2}\]

Shifting the terms we have,

\[2x + 9{x^2} - 7{x^2} + 5 = 0\]

\[2{x^2} - 2x + 5 = 0\]

Since the degree of the equation is 2, we have 2 factors or two roots.

On comparing the given equation with the standard quadratic equation\[a{x^2} + bx + c = 0\], we have\[a = 2\], \[b = - 2\] and \[c = 5\].

Now we use quadratic formula or Sridhar’s formula,

\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Substituting we have,

\[ \Rightarrow x = \dfrac{{ - ( - 2) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4(2)(5)} }}{{2(2)}}\]

\[ = \dfrac{{2 \pm \sqrt {4 - 40} }}{4}\]

\[ = \dfrac{{2 \pm \sqrt { - 36} }}{4}\]

\[ = \dfrac{{2 \pm \sqrt { - 1 \times 36} }}{4}\]

We know that \[\sqrt { - 1} = i\],

\[ = \dfrac{{2 \pm i\sqrt {36} }}{4}\]

We know that 36 is a perfect square,

\[ = \dfrac{{2 \pm 6i}}{4}\]

Taking 2 common,

\[ = \dfrac{{2(1 \pm 3i)}}{4}\]

\[ = \dfrac{{1 \pm 3i}}{2}\]

Thus we have two roots,

\[ \Rightarrow x = \dfrac{{1 + 3i}}{2}\] and \[x = \dfrac{{1 - 3i}}{2}\]. This is the required answer.

**Note:**Since we have a polynomial of degree two and hence it is called quadratic polynomial. If we have a polynomial of degree ‘n’ then we have ‘n’ roots. In the given problem we have a degree that is equal to 2. Hence the number of roots are 2. Also we know that \[\sqrt { - 1} \] is undefined and we take \[\sqrt { - 1} = i\] that is an imaginary number. Quadratic formula and Sridhar’s formula are both the same. We know that the product of two negative numbers gives us a positive number. Also keep in mind when shifting values from one side of the equation to another side of the equation, always change sign from positive to negative and vice-versa.

Recently Updated Pages

Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Which places in India experience sunrise first and class 9 social science CBSE

The list which includes subjects of national importance class 10 social science CBSE

What is pollution? How many types of pollution? Define it

State the laws of reflection of light