Question

Solve the quadratic equation
2*2-3x-5=0

Answer 1

Answer:

here is your answer. mate.

Step-by-step explanation:

If α, β are roots of equation 2x^2-3x-5=0, how would you form a Quadratic equation whose roots are α ^2, β^2?

If α, β are roots of equation 2x^2-3x-5=0, how would you form a Quadratic equation whose roots are α ^2, β^2?2α2–3α−5=02α2–3α−5=0

If α, β are roots of equation 2x^2-3x-5=0, how would you form a Quadratic equation whose roots are α ^2, β^2?2α2–3α−5=02α2–3α−5=02α2−5=3α2α2−5=3α

If α, β are roots of equation 2x^2-3x-5=0, how would you form a Quadratic equation whose roots are α ^2, β^2?2α2–3α−5=02α2–3α−5=02α2−5=3α2α2−5=3α(2α2−5)2=9α2(2α2−5)2=9α2

If α, β are roots of equation 2x^2-3x-5=0, how would you form a Quadratic equation whose roots are α ^2, β^2?2α2–3α−5=02α2–3α−5=02α2−5=3α2α2−5=3α(2α2−5)2=9α2(2α2−5)2=9α24α4–20α2+25=9α24α4–20α2+25=9α2

If α, β are roots of equation 2x^2-3x-5=0, how would you form a Quadratic equation whose roots are α ^2, β^2?2α2–3α−5=02α2–3α−5=02α2−5=3α2α2−5=3α(2α2−5)2=9α2(2α2−5)2=9α24α4–20α2+25=9α24α4–20α2+25=9α24α4–29α2+25=04α4–29α2+25=0

If α, β are roots of equation 2x^2-3x-5=0, how would you form a Quadratic equation whose roots are α ^2, β^2?2α2–3α−5=02α2–3α−5=02α2−5=3α2α2−5=3α(2α2−5)2=9α2(2α2−5)2=9α24α4–20α2+25=9α24α4–20α2+25=9α24α4–29α2+25=04α4–29α2+25=0Replacing α2α2 by xx

If α, β are roots of equation 2x^2-3x-5=0, how would you form a Quadratic equation whose roots are α ^2, β^2?2α2–3α−5=02α2–3α−5=02α2−5=3α2α2−5=3α(2α2−5)2=9α2(2α2−5)2=9α24α4–20α2+25=9α24α4–20α2+25=9α24α4–29α2+25=04α4–29α2+25=0Replacing α2α2 by xx4x2–29x+25=0