Question

Solve pls don't spam

Answer 1

Step-by-step explanation:

★ Concept :-

Here the concept of Quadratic Equation has been used. We see that we are given a quadratic equation where we need to find the zeroes of the equation and then verify the relationship between the zeroes. The best way to find the zeroes of the equation is using the method of Splitting the Middle Term. Then after that we can make groups and find the zeroes of equation. Finally we shall verify the relationship between the zeroes and coeffcients.

Let's do it !!

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★ Solution :-

Given,

» 4x² + 15x - 4x = 0

Now using the method of Splitting the middle term, we get

>> 4x² + 16x - 1x - 4 = 0

Taking the common terms, we get

>> 4x(x + 4) -1(x + 4) = 0

On grouping, we will get

>> (4x - 1)(x + 4) = 0

Here either (4x - 1) = 0 or (x + 4) = 0

So,

>> 4x - 1 = 0 or x + 4 = 0

>> 4x = 1 or x = - 4

>> x = ¼ or x = - 4

>> x = ¼ , - 4

These are the zeroes of the given equation.

$\;\underline{\boxed{\tt{Zeroes\;\:of\;\:equation,\;x\;=\;\bf{\red{\dfrac{1}{4},\;-4}}}}}$

Now,

• Let = ¼
• Let = -4
• Here a = 4
• Here b = 15
• Here c = -4

Then we know that,

→R.H.S.=

Also,

$\;\;\bf{\rightarrow\;\;R.H.S.\;=\;-\dfrac{b}{a}\;=\;\dfrac{-15}{4}}$

$\;\;\sf{\rightarrow\;\;\green{\alpha\;+\;\beta\;=\;-\dfrac{b}{a}}}$

Then here,

$\;\;\sf{\rightarrow\;\;L.H.S.\;=\;\alpha\;+\;\beta\;=\;\dfrac{1}{4}\:+\:(-4)}$

$\;\;\sf{\rightarrow\;\;L.H.S.\;=\;\alpha\;+\;\beta\;=\;\dfrac{1\:-\:16}{4}}$

$\;\;\bf{\rightarrow\;\;L.H.S.\;=\;\alpha\;+\;\beta\;=\;\dfrac{15}{4}}$

Also,

$\;\;\bf{\rightarrow\;\;R.H.S.\;=\;-\dfrac{b}{a}\;=\;\dfrac{-15}{4}}$

Clearly, L.H.S. = R.H.S. = -15/4

So, this condition is verified.

• Case II ::

$\;\;\sf{\rightarrow\;\;\blue{\alpha\;\times\;\beta\;=\;\dfrac{c}{a}}}$

Then here,

$\;\;\bf{\rightarrow\;\;L.H.S.\;=\;\alpha\;\times\;\beta\;=\;\dfrac{1}{4}\:\times\:(-4)\;=\;-1}$

Also,

$\;\;\bf{\rightarrow\;\;R.H.S.\;=\;\dfrac{c}{a}\;=\;\dfrac{-4}{4}\;=\;-1}$

Clearly, L.H.S. = R.H.S. = -1

So, this condition is satisfied.

Since both conditions are satisfied here. So the zeroes are correct.

Hence, verified.