. If the value of the expression x² – 5x + k equals to 5, for x = 0 , then the value of k is * (a) 2 (b) 0 (c) 4 (d) 5
write a pair of integers whose sum gives. c. an integer greater than both the integer....and the sum will be happen with 10,6.
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If the sum and product of the roots of the equation kx2 + 6x + 4k = 0 are real, then k =
Question: If the sum and product of the roots of the equation $k x^{2}+6 x+4 k=0$ are real, then $k=$ (a) $-\frac{3}{2}$ (b) $\frac{3}{2}$ (c) $\frac{2}{3}$ (d) $-\frac{2}{3}$
Solution:
The given quadric equation is $k x^{2}+6 x+4 k=0$, and roots are equal
Then find the value of $c$
Let $\alpha$ and $\beta$ be two roots of given equation
And, $a=k, b=6$ and,$c=4 k$
Then, as we know that sum of the roots
$\alpha+\beta=\frac{-b}{a}$
$\alpha+\beta=\frac{-6}{k}$
And the product of the roots
$\alpha \cdot \beta=\frac{c}{a}$
$\alpha \beta=\frac{4 k}{k}$
$=4$
According to question, sum of the roots = product of the roots
$\frac{-6}{k}=4$
$4 k=-6$
$k=\frac{-6}{4}$
$=\frac{-3}{2}$
Therefore, the value of $c=\frac{-3}{2}$
Thus, the correct answer is $(a)$
The given quadric equation is kx2 + 6x + 4k = 0, and roots are equal
Then find the value of c.
Let `alpha and beta`be two roots of given equation
And, a = k,b = 6 and , c = 4k
Then, as we know that sum of the roots
`alpha + beta = (-b)/a`
`alpha +beta = (-6)/a`
And the product of the roots
`alpha. beta = c/a`
`alpha beta = (4k)/k`
= 4
According to question, sum of the roots = product of the roots
`(-6)/k = 4`
`4k = -6`
`k = (-6)/4`
`= (-3)/2`
Therefore, the value of `c = (-3)/2`.
Page 2
The given quadric equation is `ax^2 + bx + c = 0`, and `sin alpha and cos beta` are roots of given equation.
And, a = a,b = b and, c = c
Then, as we know that sum of the roots
`sin alpha + cos beta - (-b)/a`…. (1)
And the product of the roots
`sin alpha .cos beta =c/a`…. (2)
Squaring both sides of equation (1) we get
`(sin alpha + cos beta)^2 = ((-b)/a)^2`
`sin^2 alpha + cos^2 beta + 2 sin alpha cos beta = b^2/a^2`
Putting the value of `sin^2 alpha + cos^2 beta = 1`, we get
`1 + 2 sin alpha cos beta = b^2/a^2`
`a^2 (1+2 sin alpha cos beta) = b^2`
Putting the value of`sin alpha.cos beta = c/a` , we get
`a^2 (1 + 2 c/a) = b^2`
`a^2 ((a+2c)/a) = b^2`
`a^2 + 2ac =b^2`
Therefore, the value of `b^2 = a^2 + 2ac`.
If the sum and product of the roots of the equation k x2+6 x+4 k=0 are real, then k=a 32b 32c 23d 23