Question

Find the zeros of the quadratic
polynomial and verify the relationship between 0 and the coefficients of x²-2x-8

Answer 1

$\huge\mathfrak{Answer:}$

Given:

• We have been given a quadratic polynomial x²-2x-8.

To Find:

• We need to find the zeroes of this polynomial and also verify the relationship between zeroes and coefficients.

Solution:

The given polynomial is x²-2x-8.

We can find the zeroes of this polynomial by the method of splitting the middle term.

We need to find two such numbers whose sum is -2 and product is -8.

Two such numbers are -4 and +2.

Substituting the value, we have

x²- 4x + 2x- 8 = 0

=> x(x - 4) + 2(x - 4)

=> (x - 4)(x + 2)

Either (x - 4) = 0 or (x + 2) = 0.

When (x - 4) = 0

=> x = 4

When (x + 2) = 0

=> x = -2

Therefore, zeroes of this polynomial are 4 and -2.

4 = α and -2 = β

Now, we need to verify the relationship between the zeroes and coefficients, we have

Sum of zeroes (α + β)

= 4 + (-2)

= 4 - 2

= 2 = (-b/a)

Product of zeroes (αβ)

= 4 × (-2)

= -8 = c/a

Hence, relationship between zeroes and coefficients is verified!!

Answer 2

$\huge\underline\mathtt{SOLUTION:-}$

$\bf{\underline{\underline \green{Given:-}}}$

• The given quadratic polynomial is x² - 2x - 8

$\bf{\underline{\underline \green{Need\: To\: Find:-}}}$

• Find the zeros of the quadratic polynomial and verify the relationship between 0 and the coefficients of x²- 2x - 8

$\bf{\underline{\underline \green{ExPlanation:-}}}$

Let p(x) = x² - 2x - 8

• Zero of the polynomial is the value of x where p(x) = 0

Putting p(x) = 0

➠ x² - 2x - 8 = 0

We find roots using splitting the middle term method.

➠ x² - 4x + 2x - 8 = 0

➠ x(x - 4) + 2(x - 4) = 0

➠ (x + 2) (x - 4) = 0

• So, x = -2, 4

$\bf{\underline{\underline \green{ThereFore:-}}}$

• $\alpha$ = -2 & $\beta$ = 4 are the zeroes of the polynomial.

p(x) = x² - 2x - 8

➠ 1x² - 2x - 8

Comparing with ax² + bx + c

$\bf{\underline{\underline \green{SO:-}}}$

• a = 1
• b = -2
• c = -8

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

$\bf{\underline{\underline \green{We\: Verify:-}}}$

Sum of zeroes = $\mathsf {\frac{Coefficient\:of\:x}{Coefficient\:of\:x^2} }$

• $\alpha$ + $\beta$ = -$\mathsf {\frac{b}{a} }$

$\bf{\underline{\underline \green{L.H.S:-}}}$

$\alpha$ + $\beta$

➠ -2 + 4

➠ 2

$\bf{\underline{\underline \green{R.H.S:-}}}$

$\mathsf {-\frac{b}{a} }$

➠ $\mathsf {-\frac{-2}{1} }$

➠ 2

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

Product of zeroes = $\mathsf {\frac{Constant\:term}{Coefficient\:of\:x^2} }$

• $\alpha$ × $\beta$ = $\mathsf {\frac{c}{a} }$

$\bf{\underline{\underline \green{L.H.S:-}}}$

༆ $\alpha$ $\beta$

➠ (-2) (4)

➠ -8

$\bf{\underline{\underline \green{R.H.S:-}}}$

༆ $\mathsf {\frac{c}{a} }$

➠ $\mathsf {\frac{-8}{1} }$

➠ -8

Since, L.H.S = R.H.S

$\bf{\underline{\underline \green{Hence:-}}}$

• $\dag$Relationship between zeros and coefficient is verified.

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$