If one root of the quadratic equation . x^2 + 6x + k = 0 is h + 2 / 6 find h and k .
Answers
ANSWER :
The value of h is -3 and value of k is -15.
Step-by-step explanation:
The given quadratic equation is
x^2+6x+k=0x
2
+6x+k=0
Quadratic formula : If a quadratic equation is ax^2+bx+c=0ax
2
+bx+c=0 , then the root of quadratic equations are
x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}x=
2a
−b±
b
2
−4ac
In the given equation a=1,b=6 adn c=k. Using quadratic formula we get
x=\dfrac{-6\pm \sqrt{6^2-4(1)(k)}}{2(1)}x=
2(1)
−6±
6
2
−4(1)(k)
x=\dfrac{-6\pm \sqrt{36-4k}}{2}x=
2
−6±
36−4k
x=\dfrac{-6\pm \sqrt{4(9-k)}}{2}x=
2
−6±
4(9−k)
x=\dfrac{-6\pm 2\sqrt{9-k}}{2}x=
2
−6±2
9−k
x=\dfrac{2(-3\pm \sqrt{9-k})}{2}x=
2
2(−3±
9−k
)
x=-3\pm \sqrt{9-k}x=−3±
9−k
The roots of the quadratic equation are x=-3+\sqrt{9-k}x=−3+
9−k
and x=-3-\sqrt{9-k}x=−3−
9−k
.
It is given that one root of quadratic equation is h + 2\sqrt6h+2
6
.
h + 2\sqrt6=-3+\sqrt{9-k}h+2
6
=−3+
9−k
h + \sqrt{4\times 6}=-3+\sqrt{9-k}h+
4×6
=−3+
9−k
h + \sqrt{24}=-3+\sqrt{9-k}h+
24
=−3+
9−k
On comparing both sides we get
h=-3h=−3
24=9-k\Rightarrow k=-1524=9−k⇒k=−15
Therefore, the value of h is -3 and value of k is -15.
#Learn more
solve the 2x²-5x+2=0 quadratic equations by formula method
https://brainly.in/question/12253224
Answer:
⇒ The given quadratic equation is 8x2−6x+k=0, comparing it with ax2+bx+c.
⇒ Then, a=8,b=−6,c=k
⇒ It is given that one root of this equation is square of the other root. So, if we assume one root to be p, the other root can be p2. So, we assume the roots to be pand p2.
⇒ Sum of the roots =a−b
∴ p+p2=8−(−6)
∴ p+p2=43
∴ 4p+4p2=3
∴ 4p2+4p−3=0 ----------- ( 1 )
⇒ Product of the roots =ac
∴ p×p2=8k
∴ p3=8k ------- ( 2 )
⇒ 4p2+4p−3=0 [ From ( 1 ) ]
⇒ 4p2−2p+6p−