4x²- 7x²+3 =0 quadratic equation

1.	4x²- 7x²+3 =0 quadratic equation
Answer» Given: 4x²- 7x²+3 =0 To Find: The ROOT of the equation Understanding the concept, How do we find it? so, firstly we factorise the QUADRACTIC equation into simple products with which make theem to linear equations so, that on further simplifing we get the roots of the quadratic equation $\\\\$ $\\$ here, As we can see that, it is a quadractic equation so, thee equation should have two solutions or two zeros. Simlarly a cubic polynomial has 3 solutions and a linear equation has 1 solution. Solution: $\longrightarrow \; \tt 4x^{2} - 7x^{2} + 3=0$ now, we subract the like terms in th eequation which are 4x^2 and 7x^2 $\\ \longrightarrow \; \tt 3x^{2} + 3 = 0$ here let's transpose the term 3 form L.H.S to R.H.S changing its singh from positive to negetive $\longrightarrow \; \tt - 3x^{2} = - 3$ Now let's transpose the NUMBER -3 which is multipliying the term x^ 2 in the left hadn SID of the equation similarly the sign here too changs from multplication to division $\longrightarrow \; \tt x^{2} = - 3 \div - 3\\ \\ \longrightarrow \; \tt x^{2} = 1 \;\;$ now lt's find the square root of 1 to fin th valu of x $\longrightarrow \; \tt x = \sqrt{1}$ case (i) $\dashrightarrow \; \tt x = 1$ case (ii) $\dashrightarrow \; \tt x = - 1$ Hence, th roots of the equation are 1 and -1 respctivly More To Know: Factorising Trinomials Of The Form ax^2 + bx + c Example ; 2x² + 7x + 5 whil factorising equations in such form we have to first multiply the first term with the last teerm in the polynomial and then chck to that w should find two numbers WHOSE sum is 7 and the product is 2 times 5 , which is the number 10 . so, the numbers are two and five now let's factorise, ↠ 2x² + 7x + 5 ↠ 2x² + 2x + 5x + 5 ↠ 2x( x + 1 ) + 5( x + 1 ) ↠ ( x + 1 ) ( 2x + 5 ) Now, let's find th roots ↠ x + 1 = 0 ↠ x = - 1 second root ↠ 2x + 5 = 0 ↠ 2x = - 5 ↠ x = - 5/2 hope this helps

Answer»

Given:

4x²- 7x²+3 =0

To Find:

The ROOT of the equation

Understanding the concept,

How do we find it?

so, firstly we factorise the QUADRACTIC equation into simple products with which make theem to linear equations so, that on further simplifing we get the roots of the quadratic equation

$\\\\$ $\\$

here,

As we can see that, it is a quadractic equation so, thee equation should have two solutions or two zeros. Simlarly a cubic polynomial has 3 solutions and a linear equation has 1 solution.

Solution:

$\longrightarrow \; \tt 4x^{2} - 7x^{2} + 3=0$

now, we subract the like terms in th eequation which are 4x^2 and 7x^2

$\\ \longrightarrow \; \tt 3x^{2} + 3 = 0$

here let's transpose the term 3 form L.H.S to R.H.S changing its singh from positive to negetive

$\longrightarrow \; \tt - 3x^{2} = - 3$

Now let's transpose the NUMBER -3 which is multipliying the term x^ 2 in the left hadn SID of the equation similarly the sign here too changs from multplication to division

$\longrightarrow \; \tt x^{2} = - 3 \div - 3\\ \\ \longrightarrow \; \tt x^{2} = 1 \;\;$

now lt's find the square root of 1 to fin th valu of x

$\longrightarrow \; \tt x = \sqrt{1}$

case (i)

$\dashrightarrow \; \tt x = 1$

case (ii)

$\dashrightarrow \; \tt x = - 1$

Hence,

th roots of the equation are 1 and -1 respctivly

More To Know:

Factorising Trinomials Of The Form ax^2 + bx + c

Example ;

2x² + 7x + 5

whil factorising equations in such form we have to first multiply the first term with the last teerm in the polynomial and then chck to that w should find two numbers WHOSE sum is 7 and the product is 2 times 5 , which is the number 10 .

so, the numbers are two and five

now let's factorise,

↠ 2x² + 7x + 5

↠ 2x² + 2x + 5x + 5

↠ 2x( x + 1 ) + 5( x + 1 )

↠ ( x + 1 ) ( 2x + 5 )

Now, let's find th roots

↠ x + 1 = 0

↠ x = - 1

second root

↠ 2x + 5 = 0

↠ 2x = - 5

↠ x = - 5/2

hope this helps