Question

Hola Brainlians. Here's a question to check your answering skill. ICSE Board : Quadratic Equations, Class 10. Give answer fast and no spam.

Solve -

i) 4x² + 15x - 4 = 0

Find the zeroes and verify the relationship with coeffcients.

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Answer 1

Answer:

α = 1/4

β = -4

Step-by-step explanation:

p(x) = 4x² + 15x - 4 = 0

Here

a = 4

b = 15

c = -4

Using Quadratic Formula or Sridharacharya Formula

x = -b ± (√b² - 4ac) / 2a

D = b² - 4ac

= (15)² - (4×4×-4)

= 225 - (-64)

= 289

Now,

√D = √289 = 17

Now

α = -b + √D /2a

= -15 + 17/2×4

= 2/8

= 1/4

β = -b-√D/2a

= -15-17/2*4

= -32/8

= -4

α = 1/4

β = -4

α+β = 1/4 - 4 = -15/4

α+β= -b/a = -15/4

Proved !

αβ= 1/4×-4= -1

αβ= c/a = 4/-4=-1

Proved !

Sorry to say I am neither Mod nor Star. :(

Answer 2

★ 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 !!

______________________________________

★ 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,

• Case I ::

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

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.