Question

find the zeros of the following quadratic polynomial and verify the relationship between the zeroes and coefficient. 5xsquare -29x+20​

Answer 1

${5x}^{2} - 29x + 20$

Factorise The Term By Middle Term Splitting.

In Middle Term Splitting Divide Middle Term Such that On Multiplying Both It should be Product of Coffecient x² and constant Term

${5x}^{2} - 29x + 20 = 0 \\ \\ {5x}^{2} - 25x - 4x + 20 = 0 \\ \\ 5x(x - 5) - 4(x - 5) = 0 \\ \\ (5x - 4)(x - 5) = 0$

Here One Zero of Polynomial

$5x - 4 = 0 \\ \\ 5x = 4 \\ \\ x = \frac{4}{5}$

Other Zero of Polynomial

$x - 5 = 0 \\ \\ x = 5$

We got Zeroes

$\alpha = \frac{4}{5} \\ \\ \beta = 5$

Polynomial

ax²+bx+c

${5x}^{2} - 29x + 20$

Coffecient of x²=5

Coffecient of x¹=-29

Coffecient of x^0=20

Let a in ax²+bx+c=5

Let B in ax²+bx+c=-29

Let c in ax²+bx+c=20

Sum Of Zeroes

$\alpha + \beta = - \frac{coffecient \: of \: x}{coffecient \: of \: {x}^{2} } \\ \\ \alpha + \beta = - \frac{b}{a} \\ \\ \frac{4}{5} + 5 = - \frac{-29}{5} \\ \\ \frac{29}{4} = + \frac{29}{4}</p><p>$

$lhs = rhs$

Product of Zeroes

$\alpha \times \beta = \frac{constant \: term}{cofficient \: of \: {x}^{2} } \\ \\ \frac{4}{5} \times 5 = \frac{20}{5} \\ \\ \frac{20}{5} = \frac{20}{5} \\ \\ lhs = rhs$

$\boxed{\mathbf{\huge{LHS=RHS}}}$

$\underline{\mathbf{\huge{Verified✓}}}$

Answer 2

Concept

A quadratic polynomial is a polynomial of the second degree with the highest degree term equal to two.

Given

The given quadratic polynomial is $5x^{2}-29x+20$.

Find

We are asked to find the zeroes of the quadratic polynomial and verify the relationship between the zeroes and the coefficient.

Solution

Firstly, factorize the given polynomial by multiplying the coefficient of the first term by the constant and then find two factors whose sum equals the coefficient of the middle term.

$5x^2-29x+20$

Splitting the middle term as $-29x=-25x-4x$

$5x^2-25x-4x+20$

Now, we will add up the first two terms by pulling out like factors and the same with the last two terms

$5x\left(x-5\right)-4\left(x-5\right)$

$\Rightarrow (x-5)(5x-4)$

Further, we will find the roots of the product by comparing it with zero

$x=5,\frac{4}{5}$

Now, we will verify the result by substituting the zeroes of the polynomial one by one in the polynomial

Firstly, substitute $x=5$ and we get

$5x^2-29x+20=5(5)^2-29(5)+20$

$\Rightarrow 5(5)^2-29(5)+20=0$

Further, substitute $x=\frac{4}{5}$, we get

$5\left(\frac{4}{5}\right)^{2}-29\left(\frac{4}{5}\right)+20=0$

Hence, the relationship between the zeroes and coefficient is verified and the zeroes of the polynomial $5x^2-29x+20$ are $x=5, \frac{4}{5}$.

#SPJ2