Question

Find the value of k when quardratic equation 7x^2+7x+k whose zeros are equal to eachother

Answer 1

Answer:

The value of k is 7/4

Step-by-step explanation:

Given the quadratic polynomial

7x² + 7x + k

Let us consider the zeroes of the polynomial be

$\alpha \: \: \: and \: \: \beta$

Again given condition is : zeroes of the polynomial are equal

$\alpha = \beta$

Therefore ,

$sum \: \: of \: the \: roots = - \frac{co \: efficient \: of \: x}{co \: efficient \: of \: {x}^{2} }$

$\alpha + \beta = - \frac{7}{7} \\ \implies \alpha + \beta = - 1 \\ \implies \alpha + \alpha = - 1 \: (since \: \alpha = \beta ) \\ \implies2 \alpha = - 1 \\ \implies \alpha = - \frac{1}{2}$

Therefore , the roots of the equation are

$\alpha = \beta = - \frac{ 1}{2}$

Now

Product of the roots=constant term/co-efficient of x²

$\alpha \beta = \frac{k}{7} \\ \implies( - \frac{1}{2} ) \times ( - \frac{1}{2} ) = \frac{k}{7} \\ \implies k = \frac{7}{4}$

Answer 2

solutions:

Let use consider the zeroes of the polynomial be

$\alpha \: and \: \beta$

Zeroes of the polynomial are equal

$\alpha = \beta$

Thus, sum of the roots = - coefficient of x / coefficient of x²

$\alpha + \beta = - 7 \div 7 \\ \alpha + \beta = - 1 \\ \alpha + \alpha = - 1 \\ 2 \alpha = - 1 \\ \alpha = - 1 \div 2 \\ \alpha = \beta = - 1 \div 2$

product of the roots = constant term / coefficient of x²

$\alpha \beta = k \div 7 \\ - 1 \div 2 \times - 1 \div 2 = k \div 7 \\ k = 7 \div 4$

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